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Version 1.3 0507

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Changes for version 1.1 included the insertion of a new chapter 5 for the 2007 specification and some minor mathematical corrections in chapters 2 and 4. Please note that these are not side-barred. Changes for version 1.2 include further corrections to chapter 4 and a correction to the numbering of the exercises in chapter 5. Changes for version 1.3 include mathematical corrections to exercises and answers.

GCE Mathematics (6360) Further Pure unit 2 (MFP2) Textbook

The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales 3644723 and a registered charity number 1073334. Registered address AQA, Devas Street, Manchester M15 6EX. Dr Michael Cresswell Director General.

Copyright ? 2007 AQA and its licensors. All rights reserved.

klmGCE Further Mathematics (6370)
Further Pure 2: Contents
Chapter 1: Complex numbers 1.1 1.2 1.3 1.4 1.5

Further Pure 2 (MFP2) Textbook

4 5 5 6 8 9 10 11 13 14 15 21 22 22 24 27 28 31 32 33 38 39 40 45 48 53 54 56 60 69 71 74 77

Introduction The general complex number The modulus and argument of a complex number The polar form of a complex number Addition, subtraction and multiplication of complex numbers of the form x + iy 1.6 The conjugate of a complex number and the division of complex numbers of the form x +iy 1.7 Products and quotients of complex numbers in their polar form 1.8 Equating real and imaginary parts 1.9 Further consideration of |z2 – z1| and arg(z2 –z1) 1.10 Loci on Argand diagrams

Chapter 2: Roots of polynomial equations 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 Introduction Quadratic equations Cubic equations Relationship between the roots of a cubic equation and its coefficients Cubic equations with related roots An important result Polynomial equations of degree n Complex roots of polynomial equations with real coefficients

Chapter 3: Summation of finite series 3.1 3.2 3.3 3.4 Introduction Summation of series by the method of differences Summation of series by the method of induction Proof by induction extended to other areas of mathematics

Chapter 4: De Moivre’s theorem and its applications 4.1 4.2 4.3 4.4 4.5 4.6 4.7 De Moivre’s theorem Using de Moivre’s theorem to evaluate powers of complex numbers Application of de Moivre’s theorem in establishing trigonometric identities Exponential form of a complex number The cube roots of unity The nth roots of unity The roots of z n = α , where α is a non-real number

continued overleaf

2

klmGCE Further Mathematics (6370)
Further Pure 2: Contents (continued)
Chapter 5: Inverse trigonometrical functions 5.1 5.2 5.3 5.4 5.5

Further Pure 2 (MFP2) Textbook

85 86 89 91 94 96 102 103 104 105 106 110 111 114 115 116 119 122 125 131 132 133 137 143

Introduction and revision The derivatives of standard inverse trigonometrical functions Application to more complex differentiation Standard integrals integrating to inverse trigonometrical functions Applications to more complex integrals

Chapter 6: Hyperbolic functions 6.1 Definitions of hyperbolic functions 6.2 Numerical values of hyperbolic functions 6.3 Graphs of hyperbolic functions 6.4 Hyperbolic identities 6.5 Osborne’s rule 6.6 Differentiation of hyperbolic functions 6.7 Integration of hyperbolic functions 6.8 Inverse hyperbolic functions 6.9 Logarithmic form of inverse hyperbolic functions 6.10 Derivatives of inverse hyperbolic functions 6.11 Integrals which integrate to inverse hyperbolic functions 6.12 Solving equations Chapter 7: Arc length and area of surface of revolution 7.1 Introduction 7.2 Arc length 7.3 Area of surface of revolution Answers to the exercises in Further Pure 2

3

klmGCE Further Mathematics (6370)
Chapter 1: Complex Numbers
1.1 1.2 1.3 1.4 1.5 1.6 Introduction The general complex number The modulus and argument of a complex number The polar form of a complex number

Further Pure 2 (MFP2) Textbook

Addition, subtraction and multiplication of complex numbers of the form x + iy The conjugate of a complex number and the division of complex numbers of the form x + iy Products and quotients of complex numbers in their polar form Equating real and imaginary parts Further consideration of z2 ? z1 and arg( z2 ? z1 )

1.7 1.8 1.9

1.10 Loci on Argand diagrams

This chapter introduces the idea of a complex number. When you have completed it, you will:
? ? ? ? ? ?

know what is meant by a complex number; know what is meant by the modulus and argument of a complex number; know how to add, subtract, multiply and divide complex numbers; know how to solve equations using real and imaginary parts; understand what an Argand diagram is; know how to sketch loci on Argand diagrams.

4

klmGCE Further Mathematics (6370)
1.1 Introduction

Further Pure 2 (MFP2) Textbook

You will have discovered by now that some problems cannot be solved in terms of real numbers. For example, if you use a calculator to evaluate ?64 you get an error message. This is because squaring every real number gives a positive value; both (+8) 2 and (?8) 2 are equal to 64. As
?1 cannot be evaluated, a symbol is used to denote it – the symbol used is i.

?1 = i
It follows that

i 2 = ?1

?64 = 64 × ?1 = 64 × ?1 = 8i.

1.2

The general complex number

The most general number that can be written down has the form x + iy , where x and y are real numbers. The term x + iy is a complex number with x being the real part and iy the imaginary part. So, both 2 + 3i and ?1 ? 4i are complex numbers. The set of real numbers, ! (with which you are familiar), is really a subset of the set of complex numbers, " . This is because real numbers are actually numbers of the form x + 0i.

5

klmGCE Further Mathematics (6370)
1.3

Further Pure 2 (MFP2) Textbook

The modulus and argument of a complex number

Just as real numbers can be represented by points on a number line, complex numbers can be represented by points in a plane. The point P(x, y) in the plane of coordinates with axes Ox and Oy represents the complex number x + iy and the number is uniquely represented by that point. The diagram of points in Cartesian coordinates representing complex numbers is called an Argand diagram. y
P(x, y) r θ O x

If the complex number x + iy is denoted by z, and hence z = x + iy, z (‘mod zed’) is defined as the distance from the origin O to the point P representing z. Thus z = OP = r. The modulus of a complex number z is given by z = x 2 + y 2 The argument of z, arg z, is defined as the angle between the line OP and the positive x-axis – usually in the range (π, –π). The argument of a complex number z is given y by arg z = θ , where tan θ = x You must be careful when x or y, or both, are negative.

6

klmGCE Further Mathematics (6370)
Example 1.3.1

Further Pure 2 (MFP2) Textbook

Find the modulus and argument of the complex number ?1 + 3i.
Solution

y P ? 1, 3

The point P representing this number, z, is shown on the diagram.
z = (?1) 2 +

(

)

3

= 2 and tan θ = 3 = ? 3. θ ?1 Therefore, arg z = 2π . O ?1 x 3 Note that when tan θ = ? 3, θ could equal + 2π or ? π . However, the sketch clearly shows 3 3 that θ lies in the second quadrant. This is why you need to be careful when evaluating the argument of a complex number.
2

( 3)

Exercise 1A

1. Find the modulus and argument of each of the following complex numbers: (a) 1 ? i, (b) 3i, (c) ?4, (d) ? 3 ? i .

Give your answers for arg z in radians to two decimal places. 2. Find the modulus and argument of each of the following complex numbers: (a) ?3 + i , (b) 3 + 4 i , (c) ?1 ? 7 i .

Give your answers for arg z in radians to two decimal places.

7

klmGCE Further Mathematics (6370)
1.4 The polar form of a complex number
In the diagram alongside, x = r cos θ and y = r sin θ . If P is the point representing the complex number z = x + iy, it follows that z may be written in the form r cos θ + ir sin θ . This is called the polar, or modulus–argument, form of a complex number.

Further Pure 2 (MFP2) Textbook y P(x, y) r θ O x

A complex number may be written in the form z = r (cosθ + i sin θ ), where z = r and arg z = θ For brevity, r (cosθ + i sin θ ) can be written as (r, θ).

Exercise 1B

1. Write the complex numbers given in Exercise 1A in polar coordinate form. 2. Find, in the form x + iy, the complex numbers given in polar coordinate form by: (a) z = 2 cos 3π + i sin 3π , 4 4

(

)

(b) 4 cos ? 2 π + i sin ? 2 π . 3 3

(

)

8

klmGCE Further Mathematics (6370)
1.5

Further Pure 2 (MFP2) Textbook

Addition, subtraction and multiplication of complex numbers of the form x + iy

Complex numbers can be subjected to arithmetic operations. Consider the example below.
Example 1.5.1

Given that z1 = 3 + 4i and z2 = 1 ? 2i, find (a) z1 + z2 , (b) z1 ? z2 and (c) z1 z2 .
Solution (a) z1 + z2 = (3 + 4i) + (1 ? 2i) = 4 + 2i. (c) z1 z2 = (3 + 4i)(1 ? 2i) (b) z1 ? z2 = (3 + 4i) ? (1 ? 2i) = 2 + 6i.

= 3 + 4i ? 6i ? 8i 2 = 3 ? 2i + 8 = 11 ? 2i. (since i 2 = ?1)

In general, if z1 = a1 + ib1 and z2 = a2 + ib2 , z1 + z2 = (a1 + a2 ) + i(b1 + b2 ) z1 ? z2 = (a1 ? a2 ) + i(b1 ? b2 ) z1 z2 = a1a2 ? b1b2 + i(a2b1 + a1b2 )

Exercise 1C

1. Find z1 + z2 and z1 z2 when: (a) z1 = 1 + 2i and z2 = 2 ? i, (b) z1 = ?2 + 6i and z2 = 1 + 2i.

9

klmGCE Further Mathematics (6370)
1.6

Further Pure 2 (MFP2) Textbook

The conjugate of a complex number and the division of complex numbers of the form x + iy

The conjugate of a complex number z = x + iy (usually denoted by z* or z ) is the complex number z* = x ? iy. Thus, the conjugate of ?3 + 2i is ?3 ? 2i, and that of 2 ? i is 2 + i. On an Argand diagram, the point representing the complex number z* is the reflection of the point representing z in the x-axis. The most important property of z* is that the product z z* is real since
z z* = ( x + iy )( x ? iy ) = x 2 + ixy ? ixy ? i 2 y 2 = x2 + y 2 .

z z* = z

2

Division of two complex numbers demands a little more care than their addition or multiplication – and usually requires the use of the complex conjugate.
Example 1.6.1
z Simplify z1 , where z1 = 3 + 4i and z2 = 1 ? 2i. 2

Solution

3 + 4i = (3 + 4i)(1 + 2i) 1 ? 2i (1 ? 2i)(1 + 2i) = 3 + 4i + 6i + 8i2 1 ? 2i + 2i ? 4i = ?5 + 10i 5 = ?1 + 2i.
Exercise 1D
2

z multiply the numerator and denominator of z1 by z2* , i.e. (1 + 2i) 2

so that the product of the denominator becomes a real number

z 1. For the sets of complex numbers z1 and z2 , find z1 where 2

(a) z1 = 4 + 2i and z2 = 2 ? i, (b) z1 = ?2 + 6i and z2 = 1 + 2i.

10

klmGCE Further Mathematics (6370)
1.7

Further Pure 2 (MFP2) Textbook

Products and quotients of complex numbers in their polar form

If two complex numbers are given in polar form they can be multiplied and divided without having to rewrite them in the form x + iy.
Example 1.7.1

Find z1 z 2 if z1 = 2 cos π + i sin π and z2 = 3 cos π ? i sin π . 6 6 3 3
Solution

(

)

(

)

z1 z2 = 2 cos π + i sin π × 3 cos π ? i sin π 3 3 6 6

( ) ( ) = 6 ( cos π cos π + i sin π cos π ? i sin π cos π ? i sin π sin π ) 3 6 3 6 3 6 3 6 cos π cos π + sin π sin π + i ( sin π cos π ? cos π sin π ) ? = 6? ? 3 6 3 6 3 6 3 6 ? ? ?
2

cos π ? π + i sin π ? π ? = 6? ? 3 6 3 6 ? ? ? = 6 cos π + i sin π . 6 6

(

(

)

)

(

)

Using the identities: cos( A ? B ) ≡ cos A cos B + sin A sin B

sin( A ? B) ≡ sin A cos B ? cos A sin B

Noting that arg z2 is ? π , it follows that the modulus of z1 z2 is the product of the modulus of 6 z1 and the modulus of z2 , and the argument of z1 z2 is the sum of the arguments of z1 and z2 .
Exercise 1E

1. (a) Find

z1 if z1 = 2 cos π + i sin π and z2 = 3 cos π ? i sin π . 3 3 6 6 z2 z (b) What can you say about the modulus and argument of z1 ? 2

(

)

(

)

11

klmGCE Further Mathematics (6370)
Example 1.7.2

Further Pure 2 (MFP2) Textbook

If z1 = ( r1 , θ1 ) and z2 = ( r2 , θ 2 ) , show that z1 z2 = r1r2 [ cos(θ1 + θ 2 ) + i sin(θ1 + θ 2 ) ] .
Solution

z1 z2 = r1 (cos θ1 + i sin θ1 ) r2 (cos θ 2 + i sin θ 2 )
2 = r1r2 ? ? cos θ1 cosθ 2 + i( sin θ1 cos θ 2 + cosθ1 sin θ 2 ) + i sin θ1 sin θ 2 ? ?

= r1r2 [ (cosθ1 cosθ 2 ? sin θ1 sin θ 2 ) + i( sin θ1 cos θ 2 + cos θ1 sin θ 2 ) ] = r1r2 [ cos(θ1 + θ 2 ) + i sin(θ1 + θ 2 ) ] .

If z1 = (r1 , θ1 ) and z2 = (r2 , θ 2 ) then z1 z2 = (r1r2 , θ1 + θ 2 ) – with the proviso that 2 π may have to be added to, or subtracted from, θ1 + θ 2 if θ1 + θ 2 is outside the permitted range for θ There is a corresponding result for division – you could try to prove it for yourself.
z ?r ? If z1 = (r1 , θ1 ) and z2 = (r2 , θ 2 ) then z1 = ? r1 , θ1 ? θ 2 ? – with the 2 2 ? ? same proviso regarding the size of the angle θ1 ? θ 2

12

klmGCE Further Mathematics (6370)
1.8 Equating real and imaginary parts

Further Pure 2 (MFP2) Textbook

z Going back to Example 1.6.1, z1 can be simplified by another method. 2

3 + 4i . Then, 1 ? 2i (1 ? 2i)(a + ib) = 3 + 4i ? a + 2b + i(b ? 2a) = 3 + 4i. Now, a and b are real and the complex number on the left hand side of the equation is equal to the complex number on the right hand side, so the real parts can be equated and the imaginary parts can also be equated: a + 2b = 3 and b ? 2a = 4.
Suppose we let a + ib =

Thus b = 2 and a = ?1, giving ?1 + 2i as the answer to a + ib as in Example 1.6.1. While this method is not as straightforward as the method used earlier, it is still a valid method. It also illustrates the concept of equating real and imaginary parts. If a + ib = c + id , where a, b, c and d are real, then a = c and b = d
Example 1.8.1

Find the complex number z satisfying the equation
(3 ? 4i) z ? (1 + i) z* = 13 + 2i.

Solution

Let z = (a + ib), then z* = (a ? ib). Thus, (3 ? 4i)(a + ib) ? (1 + i)(a ? ib) = 13 + 2i. Multiplying out, 3a ? 4ia + 3ib ? 4i 2b ? a ? ia + ib + i 2b = 13 + 2i. Simplifying, 2a + 3b + i(?5a + 4b) = 13 + 2i. Equating real and imaginary parts, 2a + 3b = 13, ?5a + 4b = 2. So, a = 2 and b = 3. Hence, z = 2 + 3i.
Exercise 1F

1. If z1 = 3, 2 π 3 (a) z1 z2 ,

(

π , find, in polar form, the complex numbers ) and z = ( 2, ? 6 )
2

z2 . z12 2. Find the complex number satisfying each of these equations:
z (b) z1 , 2

(c) z12 ,

(d) z13 ,

(e)

(a) (1 + i) z = 2 + 3i,

(b) ( z ? i)(3 + i) = 7i + 11,
13

(c) z + i = 2 z* ? 1.

klmGCE Further Mathematics (6370)
1.9

Further Pure 2 (MFP2) Textbook

Further consideration of z2 ? z1 and arg( z2 ? z1)

Section 1.5 considered simple cases of the sums and differences of complex numbers. Consider now the complex number z = z2 ? z1 , where z1 = x1 + iy1 and z2 = x2 + iy2 . The points A and B represent z1 and z2 , respectively, on an Argand diagram.

y

A ( x1 , y1 )

B ( x2 , y 2 )

C

O

x

Then z = z2 ? z1 = ( x2 ? x1 ) + i( y2 ? y1 ) and is represented by the point C ( x2 ? x1 , y2 ? y1 ). This makes OABC a parallelogram. From this it follows that
2 2 z2 ? z1 = OC = ? ? ( x2 ? x1 ) + ( y2 ? y1 ) ? ? , 1 2

that is to say z2 ? z1 is the length AB in the Argand diagram. Similarly arg( z2 ? z1 ) is the angle between OC and the positive direction of the x-axis. This in turn is the angle between AB and the positive x direction. If the complex number z1 is represented by the point A, and the complex number z2 is represented by the point B in an Argand diagram, then
z2 ? z1 = AB, and arg( z2 ? z1 ) is the angle between AB and the positive



direction of the x-axis
Exercise 1G

1. Find z2 ? z1 and arg( z2 ? z1 ) in (a) z1 = 2 + 3i, z2 = 7 + 5i, (b) z1 = 1 ? 3i, z2 = 4 + i, (c) z1 = ?1 + 2i, z2 = ?4 ? 5i.

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klmGCE Further Mathematics (6370)
1.10 Loci on Argand diagrams

Further Pure 2 (MFP2) Textbook

A locus is a path traced out by a point subjected to certain restrictions. Paths can be traced out by points representing variable complex numbers on an Argand diagram just as they can in other coordinate systems. Consider the simplest case first, when the point P represents the complex number z such that z = k . This means that the distance of P from the origin O is constant and so P will trace out a circle.
z = k represents a circle with centre O and radius k

If instead z ? z1 = k , where z1 is a fixed complex number represented by the point A on an Argand diagram, then (from Section 1.9) z ? z1 represents the distance AP and is constant. It follows that P must lie on a circle with centre A and radius k.
z ? z1 = k represents a circle with centre z1 and radius k

Note that if z ? z1 ≤ k , then the point P representing z can not only lie on the circumference of the circle, but also anywhere inside the circle. The locus of P is therefore the region on and within the circle with centre A and radius k. Now consider the locus of a point P represented by the complex number z subject to the conditions z ? z1 = z ? z2 , where z1 and z2 are fixed complex numbers represented by the points A and B on an Argand diagram. Again, using the result of Section 1.9, it follows that AP = BP because z ? z1 is the distance AP and z ? z2 is the distance BP. Hence, the locus of P is a straight line.
z ? z1 = z ? z2 represents a straight line – the perpendicular

bisector of the line joining the points z1 and z2 Note also that if z ? z1 ≤ z ? z2 the locus of z is not only the perpendicular bisector of AB, but also the whole half plane, in which A lies, bounded by this bisector. y All the loci considered so far have been related to distances – there are also simple loci in Argand diagrams involving angles. O α The simplest case is the locus of P subject to the condition that arg z = α , where α is a fixed angle.

P

x

15

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

This condition implies that the angle between OP and Ox is fixed (α ) so that the locus of P is a straight line. arg z = α represents the half line through O inclined at an angle α to the positive direction of Ox Note that the locus of P is only a half line – the other half line, shown dotted in the diagram above, would have the equation arg z = π + α , possibly ±2 π if π + α falls outside the specified range for arg z. In exactly the same way as before, the locus of a point P satisfying arg( z ? z1 ) = α , where z1 is a fixed complex number represented by the point A, is a line through A. arg( z ? z1 ) = α represents the half line through the point z1 inclined at an angle α to the positive direction of Ox
y P A O α

x

Note again that this locus is only a half line – the other half line would have the equation arg( z ? z1 ) = π + α , possibly ±2 π. Finally, consider the locus of any point P satisfying α ≤ arg( z ? z1 ) ≤ β . This indicates that the angle between AP and the positive x-axis lies between α and β , so that P can lie on or within the two half lines as shown shaded in the diagram below.
y

A O

β

α

x

16

klmGCE Further Mathematics (6370)
Exercise 1H

Further Pure 2 (MFP2) Textbook

1. Sketch on Argand diagrams the locus of points satisfying: (a) z = 3, (b) arg( z ? 1) = π , (c) z ? 2 ? i = 5. 4 2. Sketch on Argand diagrams the regions where: (a) z ? 3i ≤ 3, (b) π ≤ arg( z ? 4 ? 2i) ≤ 5π . 6 2 3. Sketch on an Argand diagram the region satisfying both z ? 1 ? i ≤ 3 and 0 ≤ arg z ≤ π . 4 4. Sketch on an Argand diagram the locus of points satisfying both z ? i = z + 1 + 2i and
z + 3i ≤ 4.

17

klmGCE Further Mathematics (6370)
Miscellaneous exercises 1

Further Pure 2 (MFP2) Textbook

1. Find the complex number which satisfies the equation 2 z + iz* = 4 ? i, where z* denotes the complex conjugate of z.
[AQA June 2001]

2. The complex number z satisfies the equation 3 ( z ? 1) = i ( z + 1) . (a) Find z in the form a + ib, where a and b are real. (b) Mark and label on an Argand diagram the points representing z and its conjugate, z * . (c) Find the values of z and z ? z* .
[NEAB March 1998]

3. The complex number z satisfies the equation zz* ? 3z ? 2 z* = 2i,
where z* denotes the complex conjugate of z. Find the two possible values of z, giving your answers in the form a + ib.
[AQA March 2000]

4. By putting z = x + iy, find the complex number z which satisfies the equation z + 2 z* = 1 + i , 2+i * where z denotes the complex conjugate of z.
[AQA Specimen]

5. (a) Sketch on an Argand diagram the circle C whose equation is z ? 3 ? i = 1. (b) Mark the point P on C at which z is a minimum. Find this minimum value. (c) Mark the point Q on C at which arg z is a maximum. Find this maximum value.
[NEAB June 1998]

18

klmGCE Further Mathematics (6370)
6. (a) Sketch on a common Argand diagram (i) the locus of points for which z ? 2 ? 3i = 3, (ii) the locus of points for which arg z = 1 π. 4 (b) Indicate, by shading, the region for which z ? 2 ? 3i < 3 and arg z < 1 π. 4
[AQA June 2001]

Further Pure 2 (MFP2) Textbook

7. The complex number z is defined by z = 1 + 3i . 1 ? 2i (a) (i) Express z in the form a + ib. (ii) Find the modulus and argument of z, giving your answer for the argument in the form pπ where ?1 < p ≤ 1. (b) The complex number z1 has modulus 2 2 and argument ? 7 π . The complex number 12 z2 is defined by z2 = z z1. (i) Show that z2 = 4 and arg z2 = π . 6 (ii) Mark on an Argand diagram the points P 1 and P 2 which represent z1 and z2 , respectively. (iii) Find, in surd form, the distance between P 1 and P 2.
[AQA June 2000]

19

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

8. (a) Indicate on an Argand diagram the region of the complex plane in which 0 ≤ arg ( z + 1) ≤ 2π . 3 (b) The complex number z is such that 0 ≤ arg ( z + 1) ≤ 2 π 3 π ≤ arg z + 3 ≤ π. and ( ) 6 (i) Sketch another Argand diagram showing the region R in which z must lie. (ii) Mark on this diagram the point A belonging to R at which z has its least possible value. (c) At the point A defined in part (b)(ii), z = z A . (i) Calculate the value of z A . (ii) Express z A in the form a + ib.
[AQA March 1999]

9. (a) The complex numbers z and w are such that z = ( 4 + 2i )( 3 ? i ) and w = 4 + 2i . 3?i

Express each of z and w in the form a + ib, where a and b are real. (b) (i) Write down the modulus and argument of each of the complex numbers 4 + 2i and 3 ? i. Give each modulus in an exact surd form and each argument in radians between ? π and π. (ii) The points O, P and Q in the complex plane represent the complex numbers 0 + 0i, 4 + 2i and 3 ? i, respectively. Find the exact length of PQ and hence, or otherwise, show that the triangle OPQ is right-angled.
[AEB June 1997]

20

klmGCE Further Mathematics (6370)
2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 Introduction Quadratic equations Cubic equations

Further Pure 2 (MFP2) Textbook

Chapter 2: Roots of Polynomial Equations

Relationship between the roots of a cubic equation and its coefficients Cubic equations with related roots An important result Polynomial equations of degree n Complex roots of polynomial equations with real coefficients

This chapter revises work already covered on roots of equations and extends those ideas. When you have completed it, you will: ? ? ? ? ? ? know how to solve any quadratic equation; know that there is a relationship between the number of real roots and form of a polynomial equation, and be able to sketch graphs; know the relationship between the roots of a cubic equation and its coefficients; be able to form cubic equations with related roots; know how to extend these results to polynomials of higher degree; know that complex conjugates are roots of polynomials with real coefficients.

21

klmGCE Further Mathematics (6370)
2.1 Introduction

Further Pure 2 (MFP2) Textbook

You should have already met the idea of a polynomial equation. A polynomial equation of degree 2, one with x 2 as the highest power of x, is called a quadratic equation. Similarly, a polynomial equation of degree 3 has x 3 as the highest power of x and is called a cubic equation; one with x 4 as the highest power of x is called a quartic equation. In this chapter you are going to study the properties of the roots of these equations and investigate methods of solving them.

2.2

Quadratic equations

You should be familiar with quadratic equations and their properties from your earlier studies of pure mathematics. However, even if this section is familiar to you it provides a suitable base from which to move on to equations of higher degree. You will know, for example, that quadratic equations of the type you have met have two roots (which may be coincident). There are normally two ways of solving a quadratic equation – by factorizing and, in cases where this is impossible, by the quadratic formula. Graphically, the roots of the equation ax 2 + bx + c = 0 are the points of intersection of the curve y = ax 2 + bx + c and the line y = 0 (i.e. the x-axis). For example, a sketch of part of y = x 2 + 2 x ? 8 is shown below.
y

(–4, 0)

(2, 0)

x

(0, –8)

The roots of this quadratic equation are those of ( x ? 2)( x + 4) = 0, which are x = 2 and ? 4.

22

klmGCE Further Mathematics (6370)
y

Further Pure 2 (MFP2) Textbook

A sketch of part of the curve y = x 2 ? 4 x + 4 is shown below.

(0, 4)

O

(2, 0)

x

In this case, the curve touches the x-axis. The equation x 2 ? 4 x + 4 = 0 may be written as ( x ? 2) 2 = 0 and x = 2, a repeated root. Not all quadratic equations are as straightforward as the ones considered so far. A sketch of part of the curve y = x 2 ? 4 x + 5 is shown below.
y

(0, 5) (2, 1)
O x

This curve does not touch the x-axis so the equation x 2 ? 4 x + 5 = 0 cannot have real roots. 2 ? ? Certainly, x 2 ? 4 x + 5 will not factorize so the quadratic formula ? x = ?b ± b ? 4ac ? has to 2a ? ?

be used to solve this equation. This leads to x = 4 ± 16 ? 20 and, using ideas from 2 4 2i ± Chapter1, this becomes or 2 ± i. It follows that the equation x 2 ? 4 x + 5 = 0 does have 2 two roots, but they are both complex numbers. In fact the two roots are complex conjugates. You may also have observed that whether a quadratic equation has real or complex roots depends on the value of the discriminant b 2 ? 4ac. The quadratic equation ax 2 + bx + c = 0 , where a, b and c are real, has complex roots if b 2 ? 4ac < 0

23

klmGCE Further Mathematics (6370)
Exercise 2A

Further Pure 2 (MFP2) Textbook

1. Solve the equations (a) x 2 + 6 x + 10 = 0, (b) x 2 + 10 x + 26 = 0.

2.3

Cubic equations

As mentioned in the introduction to this chapter, equations of the form ax 3 + bx 2 + cx + d = 0 are called cubic equations. All cubic equations have at least one real root – and this real root is not always easy to locate. The reason for this is that cubic curves are continuous – they do not have asymptotes or any other form of discontinuity. Also, as x → ∞, the term ax3 becomes the dominant part of the expression and ax3 → ∞ (if a > 0) , whilst ax3 → ?∞ when x → ?∞. Hence the curve must cross the line y = 0 at least once. If a < 0, then ax3 → ?∞ as x → ∞, and ax3 → ∞ as x → ?∞ and this does not affect the result.

24

klmGCE Further Mathematics (6370)
y

Further Pure 2 (MFP2) Textbook

A typical cubic equation, y = ax3 + bx 2 + cx + d with a > 0, can look like any of the sketches below. The equation of this curve has three real roots because the curve crosses the line y = 0 at three points.

O

x

y

y

O

x

O

x

In each of the two sketch graphs above, the curve crosses the line y = 0 just once, indicating just one real root. In both cases, the cubic equation will have two complex roots as well as the single real root.
Example 2.3.1

(a) Find the roots of the cubic equation x3 + 3x 2 ? x ? 3 = 0. (b) Sketch a graph of y = x3 + 3x 2 ? x ? 3.
Solution (a)
y
4

If f ( x) = x3 + 3x 2 ? x ? 3, then f (1) = 1 + 3 ? 1 ? 3 = 0. Therefore x ? 1 is a factor of f(x). ∴ f ( x) = ( x ? 1)( x 2 + 4 x + 3) = ( x ? 1)( x + 3)( x + 1). Hence the roots of f(x) = 0 are 1, –3 and –1.

(b)

2

-5

0

0

5

x

-2

25

klmGCE Further Mathematics (6370)
Example 2.3.2

Further Pure 2 (MFP2) Textbook

Find the roots of the cubic equation x3 + 4 x 2 + x ? 26 = 0.
Solution

Let f ( x) = x3 + 4 x 2 + x ? 26. Then f (2) = 8 + 16 + 2 ? 26 = 0. Therefore x ? 2 is a factor of f(x), and f ( x) = ( x ? 2)( x 2 + 6 x + 13). The quadratic in this expression has no simple roots, so using the quadratic formula on x 2 + 6 x + 13 = 0,
2 x = ?b ± b ? 4ac 2a = ?6 ± 36 ? 52 2 = ?6 ± 4i 2 = ?3 ± 2i.

Hence the roots of f ( x) = 0 are 2 and ?3 ± 2i.
Exercise 2B

1. Solve the equations (a) x3 ? x 2 ? 5 x ? 3 = 0, (b) x3 ? 3x 2 + 4 x ? 2 = 0, (c) x3 + 2 x 2 ? 3x ? 10 = 0.

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klmGCE Further Mathematics (6370)
2.4

Further Pure 2 (MFP2) Textbook

Relationship between the roots of a cubic equation and its coefficients

As a cubic equation has three roots, which may be real or complex, it follows that if the general cubic equation ax 3 + bx 2 + cx + d = 0 has roots α , β and γ , it may be written as a( x ? α )( x ? β )( x ? γ ) = 0. Note that the factor a is required to ensure that the coefficients of x 3 are the same, so making the equations identical. Thus, on expanding the right hand side of the identity, ax 3 + bx 2 + cx + d ≡ a ( x ? α )( x ? β )( x ? γ )
≡ ax 3 ? a(α + β + γ ) x 2 + a (αβ + βγ + γα ) x ? a αβγ .

The two sides are identical so the coefficients of x 2 and x can be compared, and also the number terms, b = ?a(α + β + γ ) c = a(αβ + βγ + γα ) d = ?a αβγ . If the cubic equation ax 3 + bx 2 + cx + d = 0 has roots α , β and γ , then b ∑α = ? a , c ∑ αβ = a , αβγ = ? d a Note that ∑ α means the sum of all the roots, and that ∑ αβ means the sum of all the possible products of roots taken two at a time.
Exercise 2C

1. Find ∑ α , ∑ αβ and αβγ for the following cubic equations: (a) x3 ? 7 x 2 + 12 x + 5 = 0, (b) 3x 3 + 4 x 2 ? 7 x + 2 = 0. 2. The roots of a cubic equation are α , β and γ . If ∑ α = 3, ∑ αβ = 7 and αβγ = ?5, state 2 the cubic equation.

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klmGCE Further Mathematics (6370)
2.5 Cubic equations with related roots

Further Pure 2 (MFP2) Textbook

The example below shows how you can find equations whose roots are related to the roots of a given equation without having to find the actual roots. Two methods are given.
Example 2.5.1

The cubic equation x3 ? 3 x 2 + 4 = 0 has roots α , β and γ . Find the cubic equations with: (a) roots 2α , 2 β and 2γ , (b) roots α ? 2, β ? 2 and γ ? 2, (c) roots 1 , 1 and 1 .

α β

γ

Solution: method 1

From the given equation,

∑α = 3 ∑ αβ = 0
αβγ = ?4.

(a) Hence

∑ 2α = 2∑ α = 6 ∑ 2α ? 2 β = 4∑ αβ = 0
2α ? 2 β ? 2γ = 8αβγ = ?32.

From which the equation of the cubic must be x3 ? 6 x 2 + 0 x + 32 = 0 or x 3 ? 6 x 2 + 32 = 0. (b)

∑ (α ? 2) = ( ∑ α ) ? 6 = 3 ? 6 = ?3 ∑ (α ? 2)( β ? 2) = ∑ αβ ? 2∑ α ? 2∑ β + (4 × 3) = ∑ αβ ? 4∑ α + 12
= 0 ? 12 + 12 = 0. (α ? 2)( β ? 2)(γ ? 2) = αβγ ? 2∑ αβ + 4∑ α ? 8 = ?4 ? 0 + 12 ? 8 = 0.

Hence the equation of the cubic must be x3 + 3x 2 = 0.

28

klmGCE Further Mathematics (6370)
(c)
1 = 1+ 1 +1 ∑α α β γ ∑αβ =

Further Pure 2 (MFP2) Textbook

αβγ

= 0 = 0. ?4 1?1 = 1 + 1 + 1 ∑

α β

αβ

βγ

γα

=

αβγ

∑α

= 3 = ? 3. 4 ?4 1 ? 1 ?1 = 1 = ?1. α β γ ?4 4

So that the cubic equation with roots 1 , 1 and 1 is

α β

γ

or

x3 ? 0 x 2 ? 3 x + 1 = 0 4 4 3 4 x ? 3 x + 1 = 0.

29

klmGCE Further Mathematics (6370)
Solution: method 2

Further Pure 2 (MFP2) Textbook

The second method of finding the cubic equations in Example 2.5.1 is shown below. It is not always possible to use this second method, but when you can it is much quicker than the first.

(a) As the roots are to be 2α , 2 β and 2γ , it follows that, if X = 2 x, then a cubic equation in X must have roots which are twice the roots of the cubic equation in x. As the equation in x is x3 ? 3x 2 + 4 = 0, if you substitute x = X the equation in X becomes 2 3 2 X ? 3 X + 4 = 0, 2 2 3 or X ? 6 X 2 + 32 = 0 as before.

( ) ( )

(b) In this case, if you put X = x ? 2 in x3 ? 3 x 2 + 4 = 0, then any root of an equation in X must be 2 less than the corresponding root of the cubic in x. Now, X = x ? 2 gives x = X + 2 and substituting into x 3 ? 3x 2 + 4 = 0 gives ( X + 2)3 ? 3( X + 2) 2 + 4 = 0 which reduces to X 3 + 3 X 2 = 0. (c) In this case you use the substitution X = 1 or x = 1 . For x 3 ? 3x 2 + 4 = 0 this gives x X 3 2 1 ? 3 1 + 4 = 0. X X 3 On multiplying by X , this gives 1? 3X + 4 X 3 = 0 or 4 X 3 ? 3X +1 = 0 as before.

( ) ( )

Exercise 2D

1. The cubic equation x3 ? x 2 ? 4 x ? 7 = 0 has roots α , β and γ . Using the first method described above, find the cubic equations whose roots are (a) 3α , 3β and 3γ , (b) α + 1, β + 1 and γ + 1, (c) 2 , 2 and 2 .

α β

γ

2. Repeat Question 1 above using the second method described above. 3. Repeat Questions 1 and 2 above for the cubic equation 2 x3 ? 3 x 2 + 6 = 0.

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klmGCE Further Mathematics (6370)
2.6 An important result
If you square α + β + γ you get (α + β + γ ) 2 = (α + β + γ )(α + β + γ )

Further Pure 2 (MFP2) Textbook

= α 2 + αβ + αγ + βα + β 2 + βγ + γα + γβ + γ 2 = α 2 + β 2 + γ 2 + 2αβ + 2 βγ + 2γα . So,

( ∑α )

2

= ∑ α 2 + 2∑ αβ , or
2

∑ α 2 = ( ∑ α ) ? 2∑ αβ for three numbers α , β and γ
This result is well worth remembering – it is frequently needed in questions involving the symmetric properties of roots of a cubic equation.
Example 2.6.1

The cubic equation x3 ? 5 x 2 + 6 x + 1 = 0 has roots α , β and γ . Find the cubic equations with (a) roots βγ , γα and αβ , (b) roots α 2 , β 2 and γ 2 . [Note that the direct approach illustrated below is the most straightforward way of solving this type of problem.]
Solution

(a)

∑α = 5 ∑ αβ = 6
αβγ = ?1 ∑ αβ ? βγ = ∑ αβ 2γ = αβγ ∑ β = αβγ ∑ α = ?1× 5 = ?5. αβ ? βγ ? γα = α 2 β 2γ 2 = (?1) 2 = +1.
Hence the cubic equation is x3 ? 6 x 2 ? 5 x ? 1 = 0.

(b)

∑ α 2 = ( ∑ α ) ? 2∑ αβ = 52 ? ( 2 × 6 ) = 13.
2

∑ α 2 β 2 = ( ∑ αβ ) ? 2∑ αβ .βγ using the same result but replacing α with αβ ,
2

β with βγ , and γ with γα .
Thus ∑ α 2 β 2 = ( ∑ αβ ) ? 2∑ αβ 2γ
2 2

= ( ∑ αβ ) ? 2αβγ ∑ α = 36 ? (2 × ?1× 5) = 46.

α 2 β 2γ 2 = (?1)2 = 1.
Hence the cubic equation is x3 ? 13x 2 + 46 x ? 1 = 0.

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klmGCE Further Mathematics (6370)
2.7 Polynomial equations of degree n

Further Pure 2 (MFP2) Textbook

The ideas covered so far on quadratic and cubic equations can be extended to equations of any degree. An equation of degree 2 has two roots, one of degree 3 has three roots – so an equation of degree n has n roots. Suppose the equation ax n + bx n ?1 + cx n ?2 + dx n ?3 … + k = 0 has n roots α , β , γ ,… then b ∑α = ? a , c ∑ αβ = a , d ∑ αβγ = ? a , (?1) n k until, finally, the product of the n roots αβγ … = . a Remember that ∑ αβ is the sum of the products of all possible pairs of roots, ∑ αβγ is the sum of the products of all possible combinations of roots taken three at a time, and so on.
In practice, you are unlikely to meet equations of degree higher than 4 so this section concludes with an example using a quartic equation.

Example 2.7.1
The quartic equation 2 x 4 + 4 x3 ? 6 x 2 + x ? 1 = 0 has roots α , β , γ and δ . Write down (a) ∑ α , (b) ∑ αβ . (c) Hence find ∑ α .
2

Solution
?4 ∑ α = α + β + γ + δ = 2 = ?2. (b) ∑ αβ = αβ + βγ + γδ + δα + αγ + βδ = ?6 = ?3. 2 2 2 (c) Now ( ∑ α ) = (α + β + γ + δ )

(a)

= α 2 + β 2 + γ 2 + δ 2 + 2(αβ + βγ + γδ + δα + αγ + βδ ) = ∑ α 2 + 2∑ αβ .

This shows that the ‘important result’ in Section 2.6 can be extended to any number of letters. Hence ∑ α 2 = ( ∑ α ) ? 2∑ αβ
2

= (?2) 2 ? 2(?3) = 10.

32

klmGCE Further Mathematics (6370)
Exercise 2E

Further Pure 2 (MFP2) Textbook

1. The quartic equation 2 x 4 ? 3x 2 + 5 x ? 8 = 0 has roots α , β , γ and δ .

β γ (a) Find the equation with roots α , , and δ . 2 2 2 2 2 (b) Find ∑ α .

2.8

Complex roots of polynomial equations with real coefficients

Consider the polynomial equation f ( x) = ax n + bx n ?1 + cx n ? 2 … + k . Using the ideas from Chapter 1, if p and q are real, f ( p + iq) = a( p + iq) n + b( p + iq ) n ?1 + … + k = u + iv , where u and v are real. n f ( p ? iq ) = a ( p ? iq ) + b( p ? iq )n ?1 + … + k Now, = u ? iv since ?i raised to an even power is real and is the same as +i raised to an even power, making the real part of f ( p ? iq ) the same as the real part of f ( p + iq). But ?i raised to an odd power is the same as +i raised to an odd power multiplied by ?1 , and odd powers of i comprise the imaginary part of f ( p ? iq ). Thus, the imaginary part of f ( p ? iq ) is ?1 times the imaginary part of f ( p + iq ). Now if p + iq is a root of f ( x) = 0, it follows that u + iv = 0 and so u = 0 and v = 0. Hence, u ? iv = 0 making f ( p ? iq ) = 0 and p ? iq a root of f ( x) = 0. If a polynomial equation has real coefficients and if p + iq, where p and q are real, is a root of the polynomial, then its complex conjugate, p ? iq, is also a root of the equation It is very important to note that the coefficients in f ( x) = 0 must be real. If f ( x) = 0 has complex coefficients, this result does not apply.

33

klmGCE Further Mathematics (6370)
Example 2.8.1

Further Pure 2 (MFP2) Textbook

The cubic equation x3 ? 3 x 2 + x + k = 0, where k is real, has one root equal to 2 ? i. Find the other two roots and the value of k.

Solution
As the coefficients of the cubic equation are real, it follows that 2 + i is also a root. Considering the sum of the roots of the equation, if γ is the third root,

(2 ? i) + (2 + i) + γ = ? ?3 = 3, 1 γ = ?1. To find k,
? k = αβγ = (2 ? i )(2 + i )(?1) = ?5, k = 5.

Example 2.8.2
The quartic equation x 4 + 2 x3 + 14 x + 15 = 0 has one root equal to 1 + 2i. Find the other three roots.

Solution
As the coefficients of the quartic are real, it follows that 1 ? 2i is also a root. Hence [ x ? (1 + 2i) ][ x ? (1 ? 2i) ] is a quadratic factor of the quartic. Now,

[ x ? (1 + 2i)][ x ? (1 ? 2i)] = x 2 ? x(1 + 2i) ? x(1 ? 2i) + (1 + 2i)(1 ? 2i)
= x 2 ? 2 x + 5.

Hence x 2 ? 2 x + 5 is a factor of x 4 + 2 x 3 + 14 x + 15. Therefore x 4 + 2 x 3 + 14 x + 15 = ( x 2 ? 2 x + 5)( x 2 + ax + b). Comparing the coefficients of x3, 2= a?2 a = 4. Considering the number terms, 15 = 3b b = 5. Hence the quartic equation may be written as ( x 2 ? 2 x + 5)( x 2 + 4 x + 3) = 0
( x 2 ? 2 x + 5)( x + 3)( x + 1) = 0,

and the four roots are 1 + 2i, 1 ? 2i, ? 3 and ? 1.

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klmGCE Further Mathematics (6370)
Exercise 2F

Further Pure 2 (MFP2) Textbook

1. A cubic equation has real coefficients. One root is 2 and another is 1 + i. Find the cubic equation in the form x 3 + ax 2 + bx + c = 0. 2. The cubic equation x3 ? 2 x 2 + 9 x ? 18 = 0 has one root equal to 3i. Find the other two roots. 3. The quartic equation 4 x 4 ? 8 x3 + 9 x 2 ? 2 x + 2 = 0 has one root equal to 1 ? i. Find the other three roots.

35

klmGCE Further Mathematics (6370)
Miscellaneous exercises 2
1. The equation
x 3 ? 3x 2 + px + 4 = 0,

Further Pure 2 (MFP2) Textbook

where p is a constant, has roots α ? β , α and α + β , where β > 0. (a) Find the values of α and β . (b) Find the value of p.
[NEAB June 1998]

2. The numbers α , β and γ satisfy the equations α 2 + β 2 + γ 2 = 22 αβ + βγ + γα = ?11. and (a) Show that α + β + γ = 0. (b) The numbers α , β and γ are also the roots of the equation
x3 + px 2 + qx + r ? 0,

where p, q and r are real. (i) Given that α = 3 + 4i and that γ is real, obtain β and γ . (ii) Calculate the product of the three roots. (iii) Write down, or determine, the values of p, q and r.
[AQA June 200]

3. The roots of the cubic equation are α , β and γ . 2 x3 + 3x + 4 = 0

(a) Write down the values of α + β + γ , αβ + βγ + γα and αβγ . (b) Find the cubic equation, with integer coefficients, having roots αβ , βγ and γα .
[AQA March 2000]

4. The roots of the equation are α , β and γ . 7 x 3 ? 8 x 2 + 23x + 30 = 0

(a) Write down the value of α + β + γ . (b) Given that 1 + 2i is a root of the equation, find the other two roots.
[AQA Specimen]

36

klmGCE Further Mathematics (6370)
5. The roots of the cubic equation
x 3 + px 2 + qx + r = 0,

Further Pure 2 (MFP2) Textbook

where p, q and r are real, are α , β and γ . (a) Given that α + β + γ = 3, write down the value of p. (b) Given also that

α 2 + β 2 + γ 2 = ?5,
(i) find the value of q, (ii) explain why the equation must have two non-real roots and one real root. (c) One of the two non-real roots of the cubic equation is 3 ? 4i. (i) Find the real root. (ii) Find the value of r.
[AQA March 1999]

6. (a) Prove that when a polynomial f ( x ) is divided by x ? a, the remainder is f ( a ) . (b) The polynomial g ( x ) is defined by
g ( x ) = 16 x 5 + px 3 + qx 2 ? 12 x ? 1,

where p and q are real constants. When g ( x ) is divided by x ? i, where i = ?1, the remainder is 3. (i) Find the values of p and q. (ii) Show that when g ( x ) is divided by 2 x ? i, the remainder is ?6i.
[AQA June 1999]

37

klmGCE Further Mathematics (6370)
3.1 3.2 3.3 3.4 Introduction Summation of series by the method of differences Summation of series by the method of induction

Further Pure 2 (MFP2) Textbook

Chapter 3: Summation of Finite Series

Proof by induction extended to other areas of mathematics

This chapter extends the idea of summation of simple series, with which you are familiar from earlier studies, to other kinds of series. When you have completed it, you will: ? ? ? ? know new methods of summing series; know which method is appropriate for the summation of a particular series; understand an important method known as the method of induction; be able to apply the method of induction in circumstances other than in the summation of series.

38

klmGCE Further Mathematics (6370)
3.1 Introduction

Further Pure 2 (MFP2) Textbook

You should already be familiar with the idea of a series – a series is the sum of the terms of a sequence. That is, the sum of a number of terms where the terms follow a definite pattern. For instance, the sum of an arithmetic progression is a series. In this case each term is bigger than the preceeding term by a constant number – this constant number is usually called the common difference. Thus,
2 + 5 + 8 + 11 + 14

is a series of 5 terms, in arithmetic progression, with common difference 3. The sum of a geometric progression is also a series. Instead of adding a fixed number to find the next consecutive number in the series, you multiply by a fixed number (called the common ratio). Thus,
2 + 6 + 18 + 54 + 162 + 486

is a series of 6 terms, in geometric progression, with common ratio 3. A finite series is a series with a finite number of terms. The two series above are examples of finite series.

39

klmGCE Further Mathematics (6370)
3.2

Further Pure 2 (MFP2) Textbook

Summation of series by the method of differences
1 + 1 + 1 +… + 1 . 1× 2 2 × 3 3 × 4 n ( n + 1)

Some problems require you to find the sum of a given series, for example sum the series

In others you have to show that the sum of a series is a specific number or a given expression. An example of this kind of problem is 1 = 1? 1 . show that 1 + 1 + 1 + … + 1× 2 2 × 3 3 × 4 n +1 n ( n + 1) The method of differences is usually used when the sum of the series is not given. Suppose you want to find the sum,
r =1

∑ ur , of a series
u1 + u2 + u3 + … + un

n

where the terms follow a certain pattern. The aim in the method of differences is to express 1 is the r th term of the first series the r th term, which will be a function of r (just as r ( r + 1) expressed as f ( r ) ? f ( r + 1) , or possibly f ( r + 1) ? f ( r ) , where f ( r ) is some function of r. If you can express ur in this way, it can be seen that setting r = 1 and then r = 2 gives
u1 + u2 = f (1) ? f ( 2 ) + f ( 2 ) ? f ( 3)

above), as the difference of two expressions in r of the same form. In other words, ur is

= f (1) ? f ( 3) . If this idea is extended to the whole series, then
r =1 r=2 r =3 u1 = f (1) ? f (1) u2 = f ( 2 ) ? f ( 3 ) u3 = f ( 3) ? f ( 4 )

$
r = n ?1 r=n

$
un ?1 = f ( n ? 1) ? f ( n ) un = f ( n ) ? f ( n + 1)

Now, adding these terms gives
u1 + u2 + u3 + … + un ?1 + un = f (1) ? f ( 2 ) +f ( 2 ) ? f ( 3) +f ( 3) ? f ( 4 ) +f ( 4 ) … + f ( n ? 1) ? f ( n ) +f ( n ) ? f ( n + 1) . The left hand side of this expression is the required sum of the series,
f (1) ? f ( n + 1) as the sum of the series.
n

hand side, nearly all the terms cancel out: f ( 2 ) , f ( 3) , f ( 4 ) , … , f ( n ) all cancel leaving just

r =1

∑ ur .

On the right

40

klmGCE Further Mathematics (6370)
Example 3.2.1
Find the sum of the series
1 + 1 + 1 +… + 1 . 1× 2 2 × 3 3 × 4 n ( n + 1)

Further Pure 2 (MFP2) Textbook

Solution
Clearly, this is not a familiar standard series, such as an arithmetic or geometric series. Nor is the answer given. So it seems that the method of differences can be applied. 1 . We need to try to split up u . The only As above, the r th term, ur , is given by r r ( r + 1) 1 in partial fractions. Suppose sensible way to do this is to express r ( r + 1) 1 = A+ B . r ( r + 1) r r + 1 Then, 1 = A(r + 1) + Br. Comparing the coefficients of r, A + B = 0. Comparing the constant terms, A = 1. Hence B = ?1 and 1 ur = =1? 1 . r ( r + 1) r r + 1 Hence, in this case the f ( r ) mentioned previously would be 1 , with f ( r + 1) = 1 , and so r r +1 on. Now, writing down the series term by term,

r =1 r=2 r =3 $ r = n ?1 r=n

1 =1? 1 =1? 1 1× 2 1 1 + 1 1 2 1 =1? 1 = 1?1 2× 3 2 2 +1 2 3 1 =1? 1 =1?1 3× 4 3 3 +1 3 4 $ $ $ 1 1 = 1 ? = 1 ?1 (n ? 1)n n ? 1 (n ? 1) + 1 n ? 1 n 1 =1? 1 n(n + 1) n n +1

Adding the columns, the left hand side becomes

r =1

∑ r (r + 1) .

n

1

Because the 1 , 1 , etc. terms 2 3

cancel, the right hand side becomes 1 ? 1 , namely the first left hand side term and the last r n +1 right hand side term. Hence, n 1 1 ∑ r (r + 1) = 1 ? n + 1 r =1
(n + 1) ? 1 n +1 n . = n +1

=

41

klmGCE Further Mathematics (6370)
Example 3.2.2
Show that r 2 ( r + 1) ? ( r ? 1) r 2 ≡ 4r 3 . Hence find
2 2

Further Pure 2 (MFP2) Textbook

r =1

∑ r 3.

n

Solution
The left hand side of the identity has a common factor, r 2 . 2 2 2 2 r 2 ( r + 1) ? ( r ? 1) r 2 ≡ r 2 ?( r + 1) ? ( r ? 1) ? ? ? 2? 2 2 = r r + 2 r + 1 ? r ? 2r + 1 ? ? ? 2 2 ? = r2 ? ? r + 2r + 1 ? r + 2r ? 1?

(

) (

)

= r 2 ( 4r ) = 4r 3 .

Now, if f (r ) = ( r ? 1) r 2 , then
2

f (r + 1) = ( r + 1 ? 1) ( r + 1)
2

2

= r 2 ( r + 1) ,
2

so that 4r 3 is of the form f ( r + 1) ? f (r ). Listing the terms in columns, as in Example 3.2.1,

r =1 r=2 r =3 $ r = ( n ? 1) r=n

4 ×13 4 × 23 4 × 33 $
3

( ) ( ) = ( 2 × 3 ) ? (1 × 2 ) = (3 × 4 ) ? ( 2 × 3 )
= 12 × 22 ? 02 ×12
2 2 2 2 2 2 2 2

$
2 2 2

4 × ( n ? 1) = ( n ? 1) n 2 ? ( n ? 2 ) ( n ? 1) 4 × n3 = n 2 ( n + 1) ? ( n ? 1) n 2 .
2 2

Adding the columns, it can be seen that the left hand side is

(

4 × 13 + 4 × 23 + 4 × 33 + % + 4n3 = 4 ∑ n3 .
r =1
2

) (
n

) (
)

)

n

Summing the right hand side, all the terms cancel out except those shaded in the scheme above, so the sum is n 2 ( n + 1) ? 02 × 12 . Hence,
4 ∑ r 3 = n 2 ( n + 1) ? 02 ×12
2 r =1

(

(

)

= n 2 ( n + 1) .
2

Hence,

r =1

∑ r 3 = 4 n2 ( n + 1)

n

1

2

, as required.

42

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

In both Examples 3.2.1 and 3.2.2, one term on each line cancelled out with a term on the next line when the addition was done. Some series may be such that a term in one line cancels with a term on a line two rows below it.

Example 3.2.3
1 Sum the series 1 + 1 + 1 + … + . 1× 5 3 × 7 5 × 9 ( 2n ? 1)( 2n + 3)

Solution
As in Example 3.2.1, the way forward is to express

( 2n ? 1)( 2n + 3)

1

in partial fractions.

( 2r ? 1)( 2r + 3) Multiplying both sides by ( 2r ? 1)( 2r + 3) ,

Let

1

=

A + B . 2r ? 1 2r + 3

1 ≡ A ( 2r + 3) + B ( 2r ? 1) .

Comparing the coefficients of r, 2 A + 2 B = 0, so A = ? B. Comparing the constant terms, 1 = 3 A ? B. Hence A = 1 and B = ? 1 . Thus, 4 4

( 2r ? 1)( 2r + 3)
Now substitute r = 1, 2, 3,%

1

1 ?1 1 . =1 4 2r ? 1 4 2r + 3

(

) (

)

1 =1 1 ?1 1 1× 5 4 1 4 5 1 =1 1 ?1 1 r=2 3× 7 4 3 4 7 [note that nothing will cancel at this stage] r =1 r =3

() () () () () () () ( )

[note that 1 1 will cancel on the first row and the third row] 4 5 1 =1 1 ?1 1 r=4 7 × 11 4 7 4 11 $ $ $ r = ( n ? 2) r = ( n ? 1) r=n

()

1 5×9

=1 1 ?1 1 4 5 4 9

( 2n ? 5)( 2n ? 1) ( 2n ? 3)( 2n + 1) ( 2n ? 1)( 2n + 3)
43

1

1

1

( ) ( ) = 1( 1 )? 1( 1 ) 4 2n ? 3 4 2n + 1 = 1 ( 1 ) ? 1 ( 1 ). 4 2n ? 1 4 2n + 3
1 1 ?1 =1 4 2n ? 5 4 2n ? 1

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

There will be two terms left at the beginning of the series when the columns are added, 1 1 and 1 1 . Likewise, there will be two terms left at the end of the series – the right 4 1 4 3 hand part of the r = n ? 1 and r = n rows. Therefore, addition gives

()

()

1 + 1 + 1 +… + 1 1 ?1 1 =1 1 +1 1 ?1 1× 5 3 × 7 5 × 9 ( 2n ? 1)( 2n + 3) 4 1 4 3 4 2n ? 1 4 2n + 3

() () ( ) ( 1+ 1 ? 1 + 1 ? =1? 4? ? 3 ( 2n + 1 2 n + 3 ) ? ?

)

? ? ?? = 1 ? 4 ? ? 2n + 3 + 2n + 1 ? ? 4 ? 3 ? ( 2n + 1)( 2n + 3) ? ? ? ? ? ? ?? 4 ( n + 1) = 1 ?4 ?? ? ? 4 ? 3 ? ( 2n + 1)( 2n + 3) ? ? ? ? ? ? n +1 . = 1 ?? 3 ? ( 2n + 1)( 2n + 3) ? ?

Exercise 3A
1. (a) Simplify r ( r + 1) ? ( r ? 1) r. (b) Use your result to obtain
r =1

∑ r.

n

2. (a) Show that

1 1 3 ? = . r (r + 1)(r + 2) (r + 1)(r + 2)(r + 3) r (r + 1)(r + 2)(r + 3)

(b) Hence sum the series

∑ r(r + 1)(r + 2)(r + 3) .
r =1

n

1

3. (a) Show that ( r + 1) ? ( r ? 1) = 6r 2 + 2.
3 3

(b) Deduce

r =1

∑ r 2.

n

44

klmGCE Further Mathematics (6370)
3.3

Further Pure 2 (MFP2) Textbook

Summation of a series by the method of induction

The method of induction is a method of summing a series of, say, n terms when the sum is given in terms of n. Suppose you have to show that the sum of n terms of a series is S(n). If you assume that the summation is true for one particular integer, say k, where k < n, then you are assuming that the sum of the first k terms is S(k). You may think that this rather begs the question but it must be understood that the result is assumed to be true for only one value of n, namely n = k . You then use this assumption to prove that the sum of the series to k + 1 terms is S ( k + 1) – that is to say that by adding one extra term, the next term in the series, the sum has exactly the same form as S(n) but with n replaced by k + 1. Finally, it is demonstrated that the result is true for n = 1. To summarise: 1 Assume that the result of the summation is true for n = k and prove that it is true for n = k + 1 2 Prove that the result is true for n = 1 Statement 1 shows that, by putting k = 1 (which is known to be true from Statement 2), the result must be true for n = 2; and Statement 1 shows that by putting k = 2 the result must be true for n = 3; and so on. By building up the result, it can be said that the summation result is true for all positive integers n. There is a formal way of writing out the method of induction which is shown in the examples below. For convenience, and comparison, the examples worked in Section 3.2 are used again here.

Example 3.3.1
Show that

r =1

∑r

n

1 = n . ( r + 1) n + 1

Solution
Assume that the result is true for n = k ; that is to say 1 + 1 + 1 +… + 1 = k . 1× 2 2 × 3 3 × 4 k ( k + 1) k + 1 Adding the next term to both sides, 1 + 1 + 1 +… + 1 1 1 . + = k + 1× 2 2 × 3 3 × 4 k ( k + 1) ( k + 1)( k + 2 ) k + 1 ( k + 1)( k + 2 ) Then k +1 1 k 1 ∑ r r +1 = k +1 + k +1 k + 2 ( ) ( )( ) r =1
= k ( k + 2) + 1 ( k + 1)( k + 2 )

2 = k + 2k + 1 ( k + 1)( k + 2 )

45

klmGCE Further Mathematics (6370)
=

Further Pure 2 (MFP2) Textbook

( k + 1)( k + 1) ( k + 1)( k + 2 )

= k +1 k +2 = k +1 , ( k + 1) + 1 which is of the same form but with k + 1 replacing k. Hence, if the result is true for n = k , it is true for n = k + 1. But it is true for n = 1 because the left hand side is 1 = 1 , and the 1× 2 2 1 1 = . Therefore the result is true for all positive integers by induction. right hand side is 1+1 2

Example 3.3.2
Show that

r =1

∑ r 3 = 4 n 2 ( n + 1)

n

1

2

.

Solution
Assume that the result is true for n = k , that is to say k 2 ( k + 1) .2 ∑ r3 = 1 4
k

r =1

Then, adding the next term to both sides
k +1 r =1

k 2 ( k + 1) ∑ r3 = 1 4

2

+ ( k + 1)

3

2 = 1 ( k + 1) ? k 2 + 4 ( k + 1) ? ? ? 4 2 2 ? = 1 ( k + 1) ? ? k + 4k + 4 ? 4 2 2 = 1 ( k + 1) ( k + 2 ) 4 2 2 k + 1) + 1? , = 1 ( k + 1) ? ( ? ? 4

which is of the same form but with k + 1 replacing k. Hence, if the result is true for n = k , it is true for n = k + 1. But it is true for k = 1 because the left hand side is 13 = 1, and the right hand side is 1 ×12 × 22 = 1. Therefore the result is true for all positive integers by induction. 4

46

klmGCE Further Mathematics (6370)
Exercise 3B
1. Prove the following results by the method of induction:

Further Pure 2 (MFP2) Textbook

(a) (1× 2 ) + ( 2 × 3) + ( 3 × 4 ) + … + n ( n + 1) = 1 n ( n + 1)( n + 2 ) . 3 (b) 12 + 22 + 32 + … + n 2 = 1 n ( n + 1)( 2n + 1) . 6 (c) (d)
r =1 n

∑ r ( r + 2 ) = 6 n ( n + 1)( 2n + 7 ). ∑ r × r ! = ( n + 1)!? 1.

n

1

r =1

47

klmGCE Further Mathematics (6370)
3.4

Further Pure 2 (MFP2) Textbook

Proof by induction extended to other areas of mathematics

The method of induction is certainly useful in the summation of series but it is not confined to this area of mathematics. This chapter concludes with a look at its use in three other connections – sequences, divisibility and de Moivre’s theorem for positive integers.

Example 3.4.1 – application to sequences
A sequence u1 , u2 , u3 , … is defined by u1 = 3
n +1 Prove by induction that for all n ≥ 1, un = 2 n ? 1 . 2 ?1

un +1 = 3 ? 2 un

( n ≥ 1) .

Solution
Assume that the result is true for n = k , that is to say
k +1 uk = 2 k ? 1 . 2 ?1

Then, using the relationship given, uk +1 = 3 ? 2 uk = 3? 2 ?1 2k ? 1
k +1

2

= 3? =
=

2 2k ? 1

(

) )

3 2k +1 ? 1 ? 2 2k ? 1

(

2k +1 ? 1

) (

(

2k +1 ? 1 3 × 2k +1 ? 3 ? 2k +1 + 2

)

( 2 × 2 ) ?1 =
k +1

2k +1 ? 1

2k +1 ? 1 k +2 = 2 k +1 ? 1 2 ?1
k +1 +1

( ) = 2 k +1 ? 1 . 2 ?1

which is of the same form as uk but with k + 1 replacing k. Hence, if the result is true for
1+1 n = k , it is true for n = k + 1. But when k = 1, u1 = 2 1 ? 1 = 3 as given. Therefore the 2 ?1 result is true for all positive integers n ≥ 1 by induction.

48

klmGCE Further Mathematics (6370)
Example 3.4.2 – application to divisibility

Further Pure 2 (MFP2) Textbook

Prove by induction that if n is a positive integer, 32 n + 7 is divisible by 8.

Solution
The best approach is a little different to that used so far. Assume that the result is true for n = k , in other words that
2 k +1 2 k +1 When n = k + 1 the expression is 3 ( ) + 7. Consider 3 ( ) + 7 ? 32 k + 7 , the difference

32 n + 7 is divisible by 8.

(

)

between the values when n = k and n = k + 1. This expression is equal to 32( k +1) ? 32 k = 32 k + 2 ? 32 k = 32 k × 32 ? 32 k = 32 k 32 ? 1 = 8 × 32 k . Thus, if 32 k + 7 is divisible by 8, and clearly 8 × 32 k is divisible by 8, it follows that 2 k +1 3 ( ) + 7 is also divisible by 8. In other words, if the result is true for n = k , it is true for
n = k + 1. But for n = 1, 32 + 7 = 16 and is divisible by 8. Hence, 32 n + 7 is divisible by 8 for all positive integers n by induction.

(

(

)

)

Example 3.4.3 – application to de Moivre’s theorem for positive integers.
Prove by induction that for integers n ≥ 1, ( cosθ + i sin θ ) = cos nθ + i sin nθ .
n

Solution
Assume that the result is true for n = k , that is to say

( cosθ + i sin θ )
k

k

= cos kθ + i sin kθ .

Multiplying both sides by cosθ + i sin θ ,

( cosθ + i sin θ ) ( cosθ + i sin θ ) = ( cos kθ + i sin kθ )( cosθ + i sin θ ) k +1 ( cosθ + i sin θ ) = cos kθ cosθ + i sin kθ cosθ + i sin θ cos kθ + i2 sin kθ sin θ = ( cos kθ cos θ ? sin kθ sin θ ) + i ( sin kθ cos θ + cos kθ sin θ ) = cos ( k + 1)θ + i sin ( k + 1)θ ,
which is of the same form but with k + 1 replacing k. Hence, if the result is true for n = k it is true for n = k + 1. But when k = 1,

?i 2 = ?1? ? ?

( cosθ + i sin θ )

1

= cos θ + i sin θ . Therefore the result is

true for all positive integers n by induction. 49

klmGCE Further Mathematics (6370)
Exercise 3C

Further Pure 2 (MFP2) Textbook

1. Prove the following results by the method of induction – in all examples n is a positive integer: (a) n3 ? n is divisible by 6. (b) 12n + 2 × 5n ?1 is divisible by 7. [Hint: consider f ( n + 1) ? 5f ( n ) where f ( n ) = 12n + 2 × 5n?1 ] (c) d x n = nx n ?1. [Hint: use the formula for differentiating a product] dx (d) x n ? 1 is divisible by x ? 1.

( )

50

klmGCE Further Mathematics (6370)
Miscellaneous exercises 3
1. Use the identity to show that
[AQA June 1999]

Further Pure 2 (MFP2) Textbook

1 =1? 1 r ( r + 1) r r + 1

. ∑ r ( r1+ 1) = n n +1
r =1

n

2. (a) Use the identity to show that

4r 3 = r 2 ( r + 1) ? ( r ? 1) r 2
2

∑ 4r
r =1 n r =1

n

3

= n 2 ( n + 1) .
2

(b) Hence find

∑ 2r ( 2r + 1) ,
2

giving your answer as a product of three factors in terms of n.
[AQA June 2000]

3. Prove by induction that


r =1

n

n r × 3r ?1 = 1 + 3 ( 2n ? 1) . 4 4

[AQA March 1999]

4. Prove by induction, or otherwise, that

∑ ( r × r !) = ( n + 1)!?1.
r =1

n

[NEAB June 1998]

5. Prove by induction that for all integers n ≥ 0, 7 n + 2 is divisible by 3.
[AQA Specimen]

6. Use mathematical induction to prove that

∑ ( r ?1)(3r ? 2) = n ( n ?1)
2

n

r =1

for all positive integers n.
[AEB June 1997]

51

klmGCE Further Mathematics (6370)
7. A sequence u1 , u2 , u3 , % is defined by
u1 = 2, un +1 = 2 ? 1 , un Prove by induction that for all n ≥ 1, un = n + 1 . n
[AQA June 1999]

Further Pure 2 (MFP2) Textbook

n ≥ 1.

8. Verify the identity

2r ? 1 ? 2 r + 1 ≡ 2 . r ( r ? 1) r ( r + 1) ( r ? 1)( r + 1)

Hence, using the method of differences, prove that 2 = 3 ? 2n + 1 . ∑ ( r ?1)( r + 1) 2 n ( n + 1)
r =2 n

[AEB January 1998]

9. The function f is defined for all non-negative integers r by f ( r ) = r 2 + r ? 1. (a) Verify that f ( r ) ? f ( r ? 1) = Ar for some integer A, stating the value of A. (b) Hence, using the method of differences, prove that n + n ). ∑r = 1 2(
2 r =1 n

[AEB January 2000]

10.

For some value of the constant A, 3r ? 2 3n + 2 . = A? ∑ r (r + 1)( r + 2 ) ( n + 1)( n + 2 )
r =1 n

(a) By setting n = 1, or otherwise, determine the value of A. (b) Use mathematical induction to prove the result for all positive integers n. (c) Deduce the sum of the infinite series 1 + 4 + 7 +% + 3n ? 2 +% . 1× 2 × 3 2 × 3 × 4 3 × 4 × 5 n ( n + 1)( n + 2 )
[AEB June 2000]

52

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

Chapter 4: De Moivre’s Theorem and its Applications
4.1 4.2 4.3 4.4 4.5 4.6 4.7 De Moivre’s theorem Using de Moivre’s theorem to evaluate powers of complex numbers Application of de Moivre’s theorem in establishing trigonometric identities Exponential form of a complex number The cube roots of unity The nth roots of unity The roots of z n = α , where α is a non-real number

This chapter introduces de Moivre’s theorem and many of its applications. When you have completed it, you will: ? ? ? ? ? ? know the basic theorem; be able to find shorter ways of working out powers of complex numbers; discover alternative methods for establishing some trigonometric identities; know a new way of expressing complex numbers; know how to work out the nth roots of unity and, in particular, the cube roots; be able to solve certain types of polynomial equations.

53

klmGCE Further Mathematics (6370)
4.1 De Moivre’s theorem

Further Pure 2 (MFP2) Textbook

In Chapter 3 (section 3.4), you saw a very important result known as de Moivre’s theorem. It was proved by induction that, if n is a positive integer, then

( cosθ + i sin θ )

n

= cos nθ + i sin nθ .

De Moivre’s theorem holds not only when n is a positive integer, but also when it is negative and even when it is fractional. Let n be a negative integer and suppose n = ? k . Then k is a positive integer and

( cosθ + i sin θ )n = ( cosθ + i sin θ )? k
= = 1 ( cosθ + i sin θ )k 1 . cos kθ + i sin kθ

Some of the results obtained in Chapter 1 can now be put to use. In order to remove i from the denominator of the expression above, the numerator and denominator are multiplied by the complex conjugate of the denominator, in this case cos kθ ? i sin kθ . Thus,
1 1 = × cos kθ ? i sin kθ cos kθ + i sin kθ cos kθ + i sin kθ cos kθ ? i sin kθ cos kθ ? i sin kθ = 2 cos kθ + i sin kθ cos kθ ? i sin kθ cos kθ ? i 2 sin 2 kθ kθ ? i sin kθ = cos 2 cos kθ + sin 2 kθ = cos kθ ? i sin kθ = cos ( ? kθ ) + i sin ( ? kθ ) = cos nθ + i sin nθ , as required.

If n is a fraction, say

p where p and q are integers, then q
q

p p ? pθ pθ ? ? cos q θ + i sin q θ ? = cos q q + i sin q q ? ? = cos pθ + i sin pθ .

[q is an integer]

But p is also an integer and so
cos pθ + i sin pθ = ( cos θ + i sin θ ) .
p

Taking the q th root of both sides,
p p cos θ + i sin θ = ( cos θ + i sin θ ) q . q q
p

54

klmGCE Further Mathematics (6370)
( cosθ + i sin θ )
p q

Further Pure 2 (MFP2) Textbook

p p It is important to point out at this stage that cos θ + i sin θ is just one value of q q

. A simple example will illustrate this. If θ = π, p = 1 and q = 2, then

( cos π + i sin π )
But ( cos π + i sin π ) = ?1
1 1 2

1 2

1 = cos 1 2 π + i sin 2 π

= i.

( cos π = ?1 and sin π = 0 )

and

?1 = ± i. So i is only one value
p

of ( cos π + i sin π ) 2 . There are, in fact, q different values of ( cos π + i sin π ) q and this will be shown in section 4.6.

( cosθ + i sin θ )

n

= cos nθ + i sin nθ

for positive and negative integers, and fractional values of n

55

klmGCE Further Mathematics (6370)
4.2
n

Further Pure 2 (MFP2) Textbook

Using de Moivre’s theorem to evaluate powers of complex numbers

One very important application of de Moivre’s theorem is in the addition of complex numbers of the form ( a + ib ) . The method for doing this will be illustrated through examples.

Example 4.2.1
Simplify cos π + i sin π . 6 6

(

)

3

Solution
It would, of course, be possible to multiply cos π + i sin π by itself three times, but this would 6 6 be laborious and time consuming – even more so had the power been greater than 3. Instead,

(

cos π + i sin π 6 6

) = cos 36π + i sin 36π
3

= cos π + i sin π 2 2 = 0+i = i.

56

klmGCE Further Mathematics (6370)
Example 4.2.2
Find

Further Pure 2 (MFP2) Textbook

(

3 +i

)

10

in the form a + ib.

Solution
Clearly it would not be practical to multiply

(

3 + i by itself ten times. De Moivre’s

)

theorem could provide an alternative method but it can be used only for complex numbers in the form cosθ + i sin θ , and 3 + i is not in this form. A technique introduced in Chapter 1 (section 1.4) can be used to express it in polar form.# y On an Argand diagram, 3 + i is represented by the point whose Cartesian coordinates are Now, r =

(

3,1 .

)

3+i

( 3)
(

2

Thus, and

+ 12 = 2 and tan θ = 1 so that θ = π . 6 3 3 + i = 2 cos π + i sin π 6 6

r θ O
3

1

(

3 +i

)

10

= 210 cos π + i sin π 6 6 = 210 cos 10π + i sin 10π 6 6 ? ? = 1024 ? 1 ? i 3 ? 2 ? ?2 = 512 1 ? i 3 .

( (

)

)

x

10

[ note that 2 is raised to the power 10 as well]

)

(

)

57

klmGCE Further Mathematics (6370)
Example 4.2.3
Simplify cos π ? i sin π . 6 6

Further Pure 2 (MFP2) Textbook

(

)

3

Solution
De Moivre’s theorem applies only to expressions in the form cosθ + i sin θ and not cosθ ? i sin θ , so the expression to be simplified must be written in the form cos ? π + i sin ? π . 6 6

( )

( ) cos ( ? π ) + i sin ( ? π ) ? ? i sin π ) = ? (cos π ? 6 6 6 6 ? ? ? = cos ( ? 3π ) + i sin ( ? 3π ) 6 6 = cos ( ? π ) + i sin ( ? π ) 2 2
3

3

= cos π ? i sin π 2 2 = ?i. Note that it is apparent from this example that ( cosθ ? i sin θ ) = cos nθ ? i sin nθ . It is very
n

important to realise that this is a deduction from de Moivre’s theorem and it must not be quoted as the theorem.

58

klmGCE Further Mathematics (6370)
Example 4.2.4
Find

Further Pure 2 (MFP2) Textbook

( ?2 + 2 3 i )

1

3

in the form a + ib.

y

( ?2, 2 3 )
r α O θ x

Solution
The complex number ?2 + 2 3 i is represented by the point whose Cartesian coordinates are ?2, 2 3 on the Argand diagram shown here.

(

)

Hence, r =

( ?2 )

2

+ 2 3 1

(

)

2

= 16 = 4, and tan θ = ? tan α = ? 2 3 so that θ = 2π . Thus 2 3
3

( ?2 + 2 3 i )

= ?2 + 2 3 i

(

)

?3

4 cos 2π + i sin 2π ? =? ? 3 3 ? ? ? cos ?3 × 2π + i sin ?3 × 2π ? = 4?3 ? ? 3 3 ? ? ? cos ( ?2π ) + i sin ( ?2π ) ? = 1 ? ? 64 ? = 1 (1 + 0 ) 64 = 1 . 64

(

(

)

)

?3

(

)

Exercise 4A
1. Prove that ( cosθ ? i sin θ ) = cos nθ ? i sin nθ .
n

2. Express each of the following in the form a + ib : (a)

( cos 3θ + i sin 3θ )
6

5

(b) (e)

(

cos π + i sin π 5 5
4

)

10

(d) (1 + i ) (g)

( 2 ? 2i )

(c) cos π + i sin π 4 4 1 (f)

(

)

2

(1 +

3i

)

5

(

3 + 3i

)

9

59

klmGCE Further Mathematics (6370)
4.3

Further Pure 2 (MFP2) Textbook

Application of de Moivre’s theorem in establishing trigonometric identities

One way of showing how these identities can be derived is to use examples. The same principles are used whichever identity is required.

Example 4.3.1
Show that cos 3θ = 4 cos3 θ ? 3cos θ .

Solution
There are several ways of establishing this result. The expansion of cos ( A + B ) can be used to express cos 2θ in terms of cos θ setting A = θ and B = θ . Similarly, the expansion of cos ( 2θ + θ ) can be used to give cos 3θ in terms of cos θ . Using de Moivre’s theorem gives a straightforward alternative method. cos 3θ + i sin 3θ = ( cosθ + i sin θ )
3 2 3

= cos3 θ + 3cos 2 θ ( i sin θ ) + 3cosθ ( i sin θ ) + ( i sin θ ) ? using the binomial expansion of ( p + q )3 ? ? ? = cos3 θ + 3i cos 2 θ sin θ ? 3cos θ sin 2 θ ? i sin 3 θ
2 ? ? using i = ?1? ?.

Now cos 3θ is the real part of the left-hand side of the equation, and the real parts of both sides can be equated, cos 3θ = cos3 θ ? 3cos θ sin 2 θ
= cos3 θ ? 3cosθ 1 ? cos 2 θ = 4 cos3 θ ? 3cos θ .

(

)

2 2 ? ?since cos θ + sin θ = 1? ?

Note that this equation will also give sin 3θ by equating the imaginary parts of both sides of the equation.

60

klmGCE Further Mathematics (6370)
Example 4.3.2
Express tan 4θ in terms of tan θ .

Further Pure 2 (MFP2) Textbook

Solution
tan 4θ = sin 4θ so expressions for sin 4θ and cos 4θ in terms of sin θ and cos θ must be cos 4θ established to start with. Using de Moivre’s theorem, cos 4θ + i sin 4θ = ( cosθ + i sin θ )
4 2 3 4

= cos 4 θ + 4 cos3 θ ( i sin θ ) + 6 cos2 θ ( i sin θ ) + 4 cosθ ( i sin θ ) + ( i sin θ )

[ using the binomial expansion ]
= cos
4

θ + 4i cos3 θ sin θ ? 6 cos 2 θ sin 2 θ ? 4i cosθ sin 3 θ + sin 4 θ
2 ? ? using i = ?1? ?

Equating the real parts on both sides of the equation, cos 4θ = cos 4 θ ? 6 cos 2 θ sin 2 θ + sin 4 θ , and equating the imaginary parts, sin 4θ = 4 cos3 θ sin θ ? 4 cos θ sin 3 θ . Now,
tan 4θ = sin 4θ cos 4θ 3 θ sin θ ? 4 cos θ sin 3 θ . = 4 cos cos 4 θ ? 6 cos 2 θ sin 2 θ + sin 4 θ

Dividing every term by cos 4 θ gives 3 4 sin θ ? 4 sin 3 θ cos θ cos θ . tan 4θ = 2 θ sin sin 4 θ + 1? 6 cos 2 θ cos 4 θ But tan θ = sin θ so cos θ 3 tan 4θ = 4 tan θ 2? 4 tan θ . 1 ? 6 tan θ + tan 4 θ

61

klmGCE Further Mathematics (6370)
Exercise 4B
1. Express sin 3θ in terms of sin θ . 2. Express tan 3θ in terms of tan θ . 3. Express sin 5θ in terms of sin θ . 4. Show that cos 6θ = cos6 θ ? 3cos 4 θ + 3cos 2 θ ? 1.

Further Pure 2 (MFP2) Textbook

62

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

So far sin nθ , cos nθ and tan nθ have been expressed in terms of sin θ , cos θ and tan θ . De Moivre’s theorem can be used to express powers of sin θ , cos θ and tan θ in terms of sines, cosines and tangents of multiple angles. First some important results must be established. Suppose z = cos θ + sin iθ . Then ?1 z ?1 = 1 = ( cosθ + i sin θ ) z = cos ( ?θ ) + i sin ( ?θ ) So,
= cos θ ? i sin θ z = cosθ + i sin θ

( = z ).

1 = cosθ + i sin θ . z Adding, and subtracting,

z + 1 = 2 cos θ , z 1 z ? = 2i sin θ . z
If z = cos θ + i sin θ z + 1 = 2 cos θ z z ? 1 = 2i sin θ z

Also,

z n = ( cos θ + i sin θ ) = cos nθ + i sin nθ
n

z ? n = 1 = ( cos θ + i sin θ ) z = cos ( ? nθ ) + i sin ( ? nθ )
?n

= cos nθ ? i sin nθ . as before, Combining z n and 1 zn
zn + 1 = 2 cos nθ , zn zn ? 1 = 2i sin nθ . zn

If z = cos θ + i sin θ , zn + 1 = 2 cos nθ zn zn ? 1 = 2i sin nθ zn A common mistake is to omit the i in 2i sin nθ , so make a point of remembering this result carefully.

63

klmGCE Further Mathematics (6370)
Example 4.3.3
Show that cos5 θ = 1 ( cos 5θ + 5cos 3θ + 10 cosθ ) . 16

Further Pure 2 (MFP2) Textbook

Solution
Suppose Then and
z = cos θ + i sin θ . 1 z + = 2 cos θ z

( 2 cosθ )

5

= z+1 z

So

() () () () 1 1 = z + 5 z + 10 z + 10 ( 1 z ) + 5( z ) + ( z ) . ? 1 ? ? 1 ? ? 1 ? 32 cos θ = ? z + ( ) ? + ?5z + 5 ( ) ? + ?10 z + 10 ( ) ? z ? ? z ? ? z ? ? ? ? 1 ? 1 ? 1 ? ? = ? z + ( ) ? + 5 ? z + ( ) ? + 10 ? z + ( ) ? . z z z ? ? ? ? ? ?
3 1 = z5 + 5z 4 1 z + 10 z z 5 3 2

( )

5

()
5

+ 10 z 2 1 z
3

3

+ 5z 1 z
5

4

+ 1 z

5

5

5

3

3

5

5

3

3

Using the results established earlier, z5 + 1 = 2 cos 5θ , z5 z3 + 1 = 2 cos 3θ , z3 z + 1 = 2 cos θ . z Hence 32 cos 5θ = 2 cos 5θ + 5 ( 2 cos 3θ ) + 10 ( 2 cos θ )

cos5 θ = 1 ( cos 5θ + 5cos 3θ + 10 cos θ ) , 16

as required.

One very useful application of the example above would be in integrating cos5 θ .

∫ cos θ = ∫ 16 ( cos 5θ + 5cos 3θ + 10 cosθ )
5

1

sin 5θ + 5sin 3θ + 10sin θ ? + c, = 1 ? ? ? 16 ? 5 3 ?

(

)

where c is an arbitrary constant.

Example 4.3.4
(a) Show that cos3 θ sin 3 θ = 1 ( 3sin 2θ ? sin 6θ ) 32 (b) Evaluate



π 2 0

cos3 θ sin 3 θ dθ .

64

klmGCE Further Mathematics (6370)
Solution (a)

Further Pure 2 (MFP2) Textbook

( ) . ( 2sin θ ) = ( z ? 1 z)
( 2 cosθ )
3

= z+1 z

3

3

3

Multiplying these,

8cos3 θ 8i3 sin 3 θ = z + 1 z
3 3

( )( ) ?64i cos θ sin θ = ? z + 1 )( z ? 1 ) ? ( ? z z ? ? ?
3

z?1 z

3

3

1 ? 2 =? ?z ? 2 ? z ? ? = z

3

( )
2

3

?3 z

( )
2

2

? 1 ? + 3 z2 ? 1 ? ? ? 1 ? ? 2? ? 2? ? 2? ?z ? ?z ? ?z ?

( )

2

3

1 ? 1 = z 6 ? 3z 2 + 3 ? ? 2 ?? 6 ?z ? z 1 ? ? 2 1 ? 6 =? ? z ? 6 ? ? 3? z ? 2 ?. z ? ? z ? ? = 2i sin 6θ and z 2 ? 12 = 2i sin 2θ . Now z 6 ? 1 6 z z

Thus,

?64i cos3 θ sin 3 θ = 2i sin 6θ ? 3 ( 2i sin 2θ ) = 2i sin 6θ ? 6i sin 2θ .

Dividing both sides by ?64i,
cos3 θ sin 3 θ = ? 1 sin 6θ + 3 sin 2θ 32 32 = 1 ( 3sin 2θ ? sin 6θ ) , 32

as required.

(b)



π 2 0

cos3 θ sin 3 θ = 1 32



π 2 0

( 3sin 2θ ? sin 6θ ) dθ
π

2 = 1 ? ? 3cos 2θ + cos 6θ ? ? ?0 32 ? 2 6 ?

3?1? ?3+1 ? = 1 ? ? 32 ? 2 6 2 6 ? ? = 1 ×8 = 1 . 32 3 12

(

)

This section concludes with an example which uses the ideas introduced here and extends into other areas of mathematics.

65

klmGCE Further Mathematics (6370)
Example 4.3.5
(a) Show that cos 5θ = cos θ 16 cos 4 θ ? 20 cos 2 θ + 5 .

Further Pure 2 (MFP2) Textbook

(

)

(b) Show that the roots of the equation 16 x 4 ? 20 x + 5 = 0 are cos rπ for r = 1, 3, 7 and 9. 10 π 3 π 5 (c) Deduce that cos 2 cos 2 = . 10 10 16

Solution (a)
Using the ideas introduced at the beginning of this section,
cos 5θ + i sin 5θ = ( cos θ + i sin θ ) .
5

Using the binomial theorem for expansion, the right-hand side of this equation becomes
cos5 θ + 5cos4 θ ( i sin θ ) + 10 cos3 θ ( i sin θ ) + 10 cos2 θ ( i sin θ ) + 5cosθ ( i sin θ ) + ( i sin θ ) .
2 3 4 5

Not every term of this expression has to be simplified. As cos 5θ is the real part of the left-hand side of the equation, it equates to the real part of the right-hand side. The real part of the right-hand side of the equation comprises those terms with even powers if i in them, since i 2 = ?1 and is real. Thus, cos 5θ = cos5 θ + 10 cos3 θ ( i sin θ ) + 5cos θ ( i sin θ )
2 4

( ) = cos θ + 10 cos θ ( ?1 + cos θ ) + 5cos θ (1 ? cos θ )
= cos5 θ + 10 cos3 θ ? sin 2 θ + 5cos θ sin 4 θ
5 3 2 2

2

using cos θ + sin θ = 1

2

2

= cos5 θ ? 10 cos3 θ + 10 cos5 θ + 5cos θ ? 10 cos3 θ + 5cos5 θ = 16 cos5 θ ? 20 cos3 θ + 5cosθ = cosθ 16 cos 4 θ ? 20 cos 2 θ + 5 .

(

)

66

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

(b) Now when cos 5θ = 0, either cos θ = 0 or 16 cos 4 θ ? 20 cos 2 θ + 5 = 0. So, putting

x = cosθ , the roots of 16 x 4 ? 20 x + 5 = 0 are the values of cos θ for which cos 5θ = 0, provided cos θ ≠ 0. 5θ = π , 3π , 5π , 7π , 9π , 11π , 13π , % But if cos 5θ = 0, 2 2 2 2 2 2 2 so that θ = π , 3π , 5π , 7π , 9π , 11π , 13π , % . 10 10 10 10 10 10 10 Also, cos 11π is the same as cos 9π , and cos 13π is the same as cos 7π , so that, 10 10 10 10 although there is an infinite number of values of θ , there are only five distinct values of cos θ and these are cos π , cos 3π , cos 5π , cos 7π and cos 9π . 10 10 10 10 10
Now cos 5π = cos π = 0 and π is, of course, a root of cosθ = 0, so that the roots of the 10 2 2 4 equation 16 x ? 20 x + 5 = 0 are cos π , cos 3π , cos 5π , cos 7π and cos 9π . 10 10 10 10 10 The roots may be written in a slightly different way as cos 7π = cos π ? 3π 10 10 = ? cos 3π , 10 and cos 9π = cos π ? 9π 10 10 = ? cos π . 10

(

)

(

)

Thus the four roots of the quartic equation 16 x 4 ? 20 x + 5 = 0 can be written as ± cos π 10 3 π and ± cos . 10

(c) From the ideas set out in Chapter 2 (section 2.7), the product of the roots of the quartic equation 16 x 4 ? 20 x + 5 = 0 is 5 so that 16
cos π ? cos π cos 3π ? cos 3π = 5 . 10 10 10 10 16 And hence, cos 2 π cos 2 3π = 5 . 10 10 16

(

)

(

)

67

klmGCE Further Mathematics (6370)
Exercise 4C
1. If z = cos θ + i sin θ write, in terms of z: (b) cos 7θ (c) sin 6θ (a) cos 4θ 2. Prove the following results: (a) cos 4θ = 8cos 4 θ ? 8cos 2 θ + 1 (b) sin 5θ = 16sin 5 θ ? 20sin 3 θ + 5sin θ (c) sin 6θ = sin θ 32 cos 5 θ ? 32 cos3 θ + 6 cos θ

Further Pure 2 (MFP2) Textbook

(d) sin 3θ

(

)

θ (d) tan 3θ = 3 tan θ ? tan 2 1 ? 3 tan θ
3

68

klmGCE Further Mathematics (6370)
4.4 Exponential form of a complex number

Further Pure 2 (MFP2) Textbook

Both cos θ and sin θ can be expressed as an infinite series in powers of θ , provided that θ is measured in radians. These are given by
cos θ = θ ? θ + θ ? θ + ...(?1) n ?1 θ + ... 2! 4! 6! ( 2n ? 2 ) !
2 4 6 2n?2

and

sin θ = θ ? θ + θ ? θ + ...(?1) n ?1 θ + ... 3! 5! 7! ( 2n ? 1)!
3 5 7

2 n ?1

There is also a series for e x given by
2 3 4 n ?1 e x = 1 + x + x + x + x + ... + x + ... 2! 3! 4! ( n ? 1)!

If iθ is substituted for x in the series for e x , e
iθ 2 3 4 n ?1 iθ ) ( iθ ) ( iθ ) iθ ) ( ( = 1 + iθ + + + + ... + + ... 2! 3! 4! ( n ? 1)!

= 1 + iθ ? θ ? iθ + θ + ... . 2! 3! 4!
2 3 4

Regrouping,

2 4 3 ? ? e iθ = 1 ? θ + θ ? % + i ? θ ? θ + % ? , 2! 4! 3! ? ?

and, using the previous results for sin θ and cosθ ,
eiθ = cos θ + i sin θ .

It is also important to note that if z = cosθ + i sin θ , then

z n = ( cos θ + i sin θ )
= e niθ ,

n

= cos nθ + i sin nθ

and if z = r ( cosθ + i sin θ ) , then z = reiθ and z n = r n e niθ . If z = r ( cos θ + i sin θ ) , then z = reiθ and z n = r n e niθ 69

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

The form reiθ is known as the exponential form of a complex number and is clearly linked to the polar form very closely. Another result can be derived from the exponential form of a complex number: eiθ = cos θ + i sin θ . So, e ?iθ = cos ( ?θ ) + isin ( ?θ ) = cosθ ? isin θ . Adding these
eiθ + e?iθ = ( cosθ + isin θ ) + ( cosθ ? isin θ ) = 2cosθ ,
iθ ? iθ cosθ = e + e . 2

or Subtracting gives

eiθ ? e?iθ = ( cosθ + isin θ ) ? ( cosθ ? isin θ ) = 2isin θ ,
iθ ? iθ sin θ = e ? e . 2i
iθ ? iθ cos θ = e + e 2 iθ ? iθ sin θ = e ? e 2i

or

Example 4.4.1
Express 2 ? 2i in the form reiθ .

Solution
The complex number 2 ? 2i is represented by the point with the coordinates ( 2, ?2 ) on an Argand diagram. Hence, and so that
r = 22 + ( ?2 ) = 8,
2

y

θ = ? tan

?1

2 =?π, 2 4
? πi 4.

2

O r

θ

x
2

2 ? 2i= 8 e

( 2, ?2 )

Exercise 4D
1. Express the following in the form reiθ : (a) 1 + i b) 3 ? i (c) 3 + 3i (d) ?2 3 + 2i

70

klmGCE Further Mathematics (6370)
4.5 The cube roots of unity

Further Pure 2 (MFP2) Textbook

The cube roots of 1 are numbers such that when they are cubed their value is 1. They must, therefore, satisfy the equation z 3 ? 1 = 0. Clearly, one root of z 3 ? 1 is z = 1 so that z ? 1 must be a factor of z 3 ? 1. Factorising, z 3 ? 1 = ( z ? 1) z 2 + z + 1 = 0.

(

)

Now z 3 ? 1 = 0 is a cubic equation and so has three roots, one of which is z = 1. The other two come from the quadratic equation z 2 + z + 1 = 0. If one of these is denoted by w, then w satisfies z 2 + z + 1 = 0 so that w2 + w + 1 = 0. It can also be shown that if w is a root of z 3 = 1, then w2 is also a root – in fact, the other root. Substituting w2 into the left-hand side of

z 3 = 1 gives w2

( )

3

= w6 = w3

( )

2

= 12 = 1, as w3 = 1 since w is a solution of z 3 = 1.

Thus the three cube roots of 1 are 1, w and w2 , where w and w2 are non-real. Of course, w can be expressed in the form a + ib by solving z 2 + z + 1 = 0 using the quadratic formula: ?1 ± 12 ? ( 4 ×1×1) 2 = ?1 ± ?3 2 = ?1 ± i 3 . 2 It doesn’t matter whether w is labelled as ?1 + i 3 or as ?1 ? i 3 because each is the square 2 2 of the other. In other words, if w = ?1 + i 3 then 2

z=

? ? w2 = ? ?1 + i 3 ? 2 ? ? =

2

1 ? 2i 3 + i 3

( )

2

4 = 1 ? 2i 3 ? 3 4 = ?2 ? 2i 3 4 = ?1 ? i 3 . 2 If w = ?1 ? i 3 , then w2 = ?1 + i 3 . 2 2 The cube roots of unity are 1, w and w2 , where

w3 = 1
1 + w + w2 = 0 and the non-real roots are ?1 ± i 3 2 71

klmGCE Further Mathematics (6370)
Both w and w2 can be expressed in exponential form. Take w = ? 1 + i 3 ; w can be represented 2 2 by the point whose Cartesian coordinates are ? 1 3? ? ? 2 , 2 ? on an Argand diagram. ? ?

Further Pure 2 (MFP2) Textbook

(

?1, 3 2 2

)
r α
?1 2

y

3 2

θ O x

? ? From the diagram, r = ? 3 ? + 1 2 ? 2 ?

2

()

2

= 1, and θ = π ? α , where tan α =
2 2

3 1

2

= 3. Thus,

2

α = π , θ = 2π and w = e 3 3
e
? 2πi 3 .

2πi 3 .

4πi ? 2πi ? The other root is w = ? e 3 ? = e 3 and can also be written as ? ?

y

Plotting the three cube roots of unity on an Argand diagram shows three points equally spaced (at intervals of 23π ) round a circle of radius 1 as shown in the diagram alongside.

2π 3 2π 3

(1,0 )

x

Example 4.5.1
Simplify w7 + w8 , where w is a complex cube root of 1.

Solution
w7 = w6 × w = w3 w8 = w6 × w2

( ) × w = 1 × w = w ( because w = 1) , = (w ) × w =1 × w = w .
2 2
3

3

2

2

2

2

2

∴ w7 + w8 = w + w2 = ?1

( because 1 + w + w2 = 0 ) .

72

klmGCE Further Mathematics (6370)
Example 4.5.2
Show that 1 + 1 + 1 = 0. 1 + w 1 + w2 w + w2

Further Pure 2 (MFP2) Textbook

Solution
1 + w + w2 = 0 so 1 + w = ? w2 , and so on. So the denominators of the left-hand side of the equation can be replaced to simplify to 1 2 + 1 + 1 . ? w ?1 ?w Multiplying the first term of this expression by w in the numerator and denominator, and the second term by w2 similarly gives w + w2 ? 1 ? w3 ? w3 = ? w ? w2 ? 1 ( as w3 = 1)
=0

( as 1 + w + w

2

=0

).

Exercise 4E
1. If w is a complex cube root of 1, find the value of (a) w10 + w11 (b) (1 + 3w ) 1 + 3w2

(

)

(c) 1 + 3w + w2

(

)

3

73

klmGCE Further Mathematics (6370)
4.6 The nth roots of unity

Further Pure 2 (MFP2) Textbook

The equation z n = 1 clearly has at least one root, namely z = 1, but it actually has many more, most of which (if not all) are complex. In fact, if n is odd z = 1 is the only real root, but if n is even z = ?1 is also a real root because ?1 raised to an even power is +1. To find the remaining roots, the right-hand side of the equation z n = 1 has to be examined. In exponential form, 1 = e0 because e0 = cos 0 + i sin 0 = 1 + i0 = 1. But also, 1 = e2 πi because e 2 πi = cos 2π + i sin 2 π = 1 + i0 = 1. Indeed 1 = e2 kπi where k is any integer. Substituting the right-hand side of the equation z n = 1 by this term gives z n = e2 kπi . Taking the nth root of both sides gives z = e n . Different integer values of k will give rise to different roots, as shown below. k = 0 gives e0 = 1,
2 kπi

k = 1 gives e

2 πi n 4πi n

= cos 2π + i sin 2π , n n = cos 4π + i sin 4π , n n
2( n ?1) πi n

k = 2 gives e
and so on until

k = n ? 1 gives e
Thus, z = e
2 kπi n

= cos

2 ( n ? 1) π 2 ( n ? 1) π + i sin . n n
2 nπi n

n = 0, 1, 2, % , n ? 1 gives the n distinct roots of the equation z n = 1.
= e 2 πi = cos 2 π + i sin 2 π = 1,
2 πi 2 πi ×e n 2 πi = 1× e n

There are no more roots because if k is set equal to n, e which is the same root as that given by k = 0. Similarly, if k is set equal to n + 1, = the same root as that given by k = 1, and so on. The n roots of z = 1 can be illustrated on an Argand diagram. All the roots lie on the circle z = 1 because the modulus of every root is 1. Also, the amplitudes of the complex numbers representing the roots are 2π , 4π , 6 π , % , 2 ( n ? 1) π . In other words, the n n n n roots are represented by n points equally spaced around the unit circle at angles of 2π starting at n (1, 0 ) – the point representing the real root z = 1.
n
2( n +1) πi e n 2 nπi e n 2 πi ×e n

=e

=

2 πi en

which is
4πi n

e

6πi n

e

e
2π n

2 πi n

2π n

2π n 2π n

e

0

e

2 ( n ?1) πi n

The equation z n = 1 has roots

z=e

2 kπi n

k = 0, 1, 2, % , ( n ? 1)

74

klmGCE Further Mathematics (6370)
Example 4.6.1

Further Pure 2 (MFP2) Textbook

Find, in the form a + ib, the roots of the equation z 6 = 1 and illustrate these roots on an Argand diagram.

Solution
z 6 = 1 = e 2 kπi

Therefore

z=e
=e

2 kπi 6 kπi 3

k = 1, 2, 3, 4, 5.

Hence the roots are k = 0,
k = 1, k = 2, k = 3, k = 4, k = 5,

z =1 z = e 3 = cos π + i sin π = 1 + i 3 3 3 2 2 = cos 2π + i sin 2π = ? 1 + i 3 3 3 2 2 z = e πi = cos π + i sin π = ?1 z=e z=e z=e
4πi 3 5πi 3 2πi 3 πi

= cos 4π + i sin 4π = ? 1 ? i 3 3 3 2 2 = cos 5π + i sin 5π = 1 ? i 3 3 3 2 2
e
2πi 3

y

πi

To summarise, the six roots are z = ±1, z = ± 1 ± i 3 and these are illustrated on 2 2 the Argand diagram alongside.

e3

–1

1
x
4πi 3

e

e

5πi 3

Two further points are worth noting. Firstly, you may need to give the arguments of the roots between ? π and + π instead of between 0 and 2 π. In example 4.6.1, the roots would be given as z = e
kπi 3

for k = 0, ± 1, ± 2, 3. Secondly, a given equation may not involve unity – for

example, if example 4.6.1 had concerned z 6 = 64, the solution would have been written

z 6 = 64 z 6 = 26 e 2 kπi z = 2e 6 k = 0, 1, 2, 3, 4, 5 and the only difference would be that the modulus of each root would be 2 instead of 1, with the consequence that the six roots of z 6 = 64 would lie on the circle z = 2 instead of z = 1.
2 kπi

75

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

Of course, there are variations on the above results. For example, you may need to find the roots of the equation 1 + z 2 + z 3 + z 4 + z 5 = 0. This looks daunting but if you can recognise the left-hand side as a geometric progression with common ratio z, it becomes more straightforward. Summing the left-hand side of the equation, 6 1 + z 2 + z 3 + z 4 + z 5 = z 5 ? 1 = 0, z ?1 2 3 4 5 so that the five roots of 1 + z + z + z + z = 0 are five of the roots of z 6 ? 1 = 0. The root to 6 be excluded is the root z = 1 because z ? 1 is indeterminate when z = 1. So the roots of z ?1 1 + z 2 + z 3 + z 4 + z 5 = 0 are z = ± 1 ± i 3 and ? 1, when written in the form a + ib. 2 2

Exercise 4F
1. Write, in the form a + ib, the roots of: (a) z 4 = 1 (b) z 5 = 32 (c) z10 = 1. In each case, show the roots on an Argand diagram. 2. Solve the equation z 4 + z 3 + z 2 + z + 1 = 0. 3. Solve the equation 1 ? 2 z + 4 z 2 + 8 z 3 = 0. 4. By considering the roots of z 5 = 1, show that cos 2π + cos 4π + cos 6π + cos 8π = ?1. 5 5 5 5

76

klmGCE Further Mathematics (6370)
4.7

Further Pure 2 (MFP2) Textbook

The roots of zn=α where α is a non-real number

Every complex number of the form a + ib can be written in the form reiθ , where r is real and θ lies in an interval of 2 π (usually from 0 to 2π or from ? π to + π). Suppose that

α = r e iθ .
Now eiθ + 2 πi = eiθ × e 2 πi = eiθ Similarly

( using e ( because e

p+q

= e p × eq =1 .

2 πi

)

)

eiθ + 2 kπi = eiθ × e2 kπi

= eiθ also. So, z n = α = reiθ + 2 kπi and, taking the nth root of both sides,
z = r ne
1 ( iθ + 2 kπi ) n 1
i(θ + 2 kπ )

= r ne

n

k = 0,1, 2, 3, %, ( n ? 1) .

1 (θ + 2 π ) n n r e

These roots can be illustrated on an Argand diagram as before. All lie on the circle z = r n and are equally spaced around the circle at 1 iθ 2 π . When k = 0, z = r n e n and this intervals of n could be taken as the starting point for the intervals of 2 π . n
1

2π n 2π n

1 iπ n n r e

The equation z n = α , where α = reiθ , has roots z =
1 i(θ + 2 kπ ) r ne n

k = 0, 1, 2, % , ( n ? 1)

77

klmGCE Further Mathematics (6370)
Example 4.7.1
Find the three roots of the equation z 3 = 2 + 2i.

Further Pure 2 (MFP2) Textbook

Solution
First, 2 + 2i must be expressed in exponential form.
y

From the diagram alongside,
r = 22 + 22 = 8, tan θ = 1, and θ = π . 4
r θ
πi

( 2, 2 )
x

So, Hence,

2 + 2i = 8 e 4 .

z3 = 8 e

(

π + 2 kπi 4

).

Taking the cube root of each side,

z= 2e = 2e
So the roots are
k = 0, k = 1, k = 2,

( π4i +2kπi )
3

(1+8k ) πi
12

k = 0, 1, 2.
πi

z = 2 e12 z = 2 e 12
9πi

z = 2 e 12

The roots can also be written

? 7πi ? ? 12 . ? or 2 e ? ? ? 2 cos π + i sin π when k = 0, and so on. 12 12

17πi

(

)

This chapter closes with one further example of the use of the principles discussed.

Example 4.7.2
Solve the equation ( z + 1) = z 5 giving your answers in the form a + ib.
5

Solution
At first sight, it is tempting to use the binomial expansion on ( z + 1) but this generates a
5

quartic equation (the terms in z 5 cancel) which would be difficult to solve. Instead, because e2 kπi = 1, the equation can be written as

( z + 1)

5

= e 2 kπi z 5 .

78

klmGCE Further Mathematics (6370)
Taking the fifth root of each side,
2 kπi

Further Pure 2 (MFP2) Textbook

z +1 = e 5 z k = 1, 2, 3, 4. Note that k = 0 is excluded because this would give z + 1 = z, and in any case as the equation is really a quartic equation it will have only four roots.

Solving the equation for z,
2 kπi ? 1= z? ? e 5 ? 1? ? ? z = 2 kπ1 . i e 5 ?1

k = 1, 2, 3, 4

or

The next step is new to this section and is well worth remembering. The term e 5 can be written as cos 2kπ + i sin 2kπ making the denominator have the form p + iq. The numerator 5 5 and denominator of the right-hand side of the equation can then be multiplied by p ? iq to remove i from the denominator. As p would then equal cos 2kπ ? 1 and q would equal 5 2 k π sin , this would be a rather cumbersome method. Instead, the numerator and 5 denominator of the right-hand side of the equation are multiplied by e will be apparent later). Thus,
? kπi 5

2 kπi

(for reasons which

z=
e

1
2 kπi 5

?1
e

,

So

z=

? kπi 5

πi ? kπi ? e 2k ? ? kπi ? 5 e 5 ??e 5 ? ?

=
e
iθ ? iθ But e ? e = sin θ so that e 2i
kπi 5

e
? kπi 5

? kπi 5

?e

? kπi 5

.

?e

? kπi 5

= 2i sin kπ and so, 5
z=
e
? kπi 5

cos ? kπ + i sin ? kπ 5 5 = 2i sin kπ 5

( )

2i sin kπ 5

( )

( ) = ? 1 ? 1 i cot ( kπ ) 2 2 5
= 1 cot kπ ? 1 2i 5 2

k = 1, 2,3, 4 as required

79

klmGCE Further Mathematics (6370)
Exercise 4G
1. Solve the following equations: (b) z 3 = 1 ? i (a) z 4 = 16i (d) z 2 = ?1 (e)

Further Pure 2 (MFP2) Textbook

(c) z 8 = 1 ? 3 i (f)

( z + 1)

3

= 8i

( z ? 1)

5

= z5

80

klmGCE Further Mathematics (6370)
Miscellaneous exercises 4

Further Pure 2 (MFP2) Textbook

1. (a) Write down the modulus and argument of the complex number ?64. (b) Hence solve the equation z 4 + 64 = 0 giving your answers in the form r ( cos θ + i sin θ ) , where r > 0 and ? π < θ ≤ π. (c) Express each of these four roots in the form a + ib and show, with the aid of a diagram, that the points in the complex plane which represent them form the vertices of a square.
[AEB June 1996]

2. (a) Express each of the complex numbers 1+ i and 3 ?i in the form r ( cos θ + i sin θ ) , where r > 0 and ? π < θ ≤ π. (b) Using your answers to part (a), (i) show that (ii) solve the equation

(

3 ?i

)

5

(1 + i )

10

= ? 1 + 3 i, 2 2

z 3 = (1 + i )

(

3 ?i

)

giving your answers in the form a + ib, where a and b are real numbers to be determined to two decimal places.
[AQA June 2001]

3. (a) By considering z = cos θ + i sin θ and using de Moivre’s theorem, show that sin 5θ ≡ sin θ 16sin 4 θ ? 20sin 2 θ + 5 . (b) Find the exact values of the solutions of the equation 16 x 4 ? 20 x 2 + 5 = 0. (c) Deduce the exact values of sin π and sin 2π , explaining clearly the reasons for your 5 5 answers.
[AQA January 2002]

(

)

81

klmGCE Further Mathematics (6370)
z 2 + z + 1 = 0.

Further Pure 2 (MFP2) Textbook

4. (a) Show that the non-real cube roots of unity satisfy the equation

(b) The real number a satisfies the equation 1 1 + = 1, 2 2 2 a ?ω +ω a +ω ?ω where ω is one of the non-real cube roots of unity. Find the possible values of a.
[AQA June 2000]

5. (a) Verify that is a root of the equation

z1 = 1 + e 5

πi

( z ? 1)

5

= ?1.

(b) Find the other four roots of the equation. (c) Mark on an Argand diagram the points corresponding to the five roots of the equation. Show that these roots lie on a circle, and state the centre and radius of the circle. (d) By considering the Argand diagram, find (i) arg z1 in terms of π, (ii) z1 in the form a cos π , where a and b are integers to be determined. b
[AQA Specimen]

6. (a) (i) Show that w =

2 πi e5

is one of the fifth roots of unity.

(ii) Show that the other fifth roots of unity are 1, w2 , w3 and w4 . (b) Let p = w + w4 and q = w2 + w3 , where w = e (i) Show that p + q = ?1 and pq = ?1. (ii) Write down the quadratic equation, with integer coefficients, whose roots are p and q. (iii) Express p and q as integer multiples of cos 2π and cos 4π , respectively, 5 5 (iv) Hence obtain the values of cos 2π and cos 4π in surd form. 5 5
[NEAB June 1998]
2 πi 5 .

82

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

7. (a) (i) Use de Moivre’s theorem to show that if z = cos θ + i sin θ , then zn + 1 = 2 cos nθ . zn (ii) Write down the corresponding result for z n ? 1 . zn (b) (i) Show that

( ) ( ) = A ??? z ? z1 ??? + B ??? z ? z1 ??? ,
z+1 z
3

z?1 z

3

6

2

6

2

where A and B are numbers to be determined. (ii) By substituting z = cos θ + i sin θ in the above identity, deduce that cos3 θ sin 3 θ = 1 ( 3sin 2θ ? sin 6θ ) . 32
[AQA June 2000]

8. (a) (i) Express e 2 ? e



? iθ 2

in terms of sin θ . 2

(ii) Hence, or otherwise, show that 1 = ? 1 ? i cot θ , iθ 2 2 2 e ?1

( e θ ≠ 1) .
i

(b) Derive expressions, in the form eiθ where ? π < θ ≤ π, for the four non-real roots of the equation z 6 = 1.

(c) The equation

( ) =1
w +1 w
6

( *)

has one real root and four non-real roots. (i) Explain why the equation has only five roots in all. (ii) Find the real root. (iii) Show that the non-real roots are 1 , z1 ? 1 1 , z2 ? 1 1 , z3 ? 1 1 , z4 ? 1

where z1 , z2 , z3 and z4 are the non-real roots of the equation z 6 = 1. (iv) Deduce that the points in an Argand diagram that represents the roots of equation (*) lie on a straight line.
[AQA March 2000]

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klmGCE Further Mathematics (6370)
(b) Show that one of the roots of the equation z 3 = 2 + 2i

Further Pure 2 (MFP2) Textbook

9. (a) Express the complex number 2 + 2i in the form reiθ , where r > 0 and ? π < θ ≤ π.

is 2 e12 , and find the other two roots giving your answers in the form reiθ , where r is a surd and ? π < θ ≤ π. (c) Indicate on an Argand diagram points A, B and C corresponding to the three roots found in part (b). (d) Find the area of the triangle ABC, giving your answer in surd form. (e) The point P lies on the circle through A, B and C. Denoting by w, α , β and γ the complex numbers represented by P, A, B and C, respectively, show that

πi

( w ? α )2 + ( w ? β )2 + ( w ? γ ) 2
[AQA June 1999]

= 6.

84

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

Chapter 5: Inverse Trigonometrical Functions
5.1 5.2 5.3 5.4 5.5 Introduction and revision The derivatives of standard inverse trigonometrical functions Applications to more complex differentiation Standard integrals integrating to inverse trigonometrical functions Applications to more complex integrals

This chapter revises and extends work on inverse trigonometrical functions. When you have studied it, you will: ? ? ? ? be able to recognise the derivatives of standard inverse trigonometrical functions; be able to extend techniques already familiar to you to differentiate more complicated expressions; be able to recognise algebraic expressions which integrate to standard integrals; be able to rewrite more complicated expressions in a form that can be reduced to standard integrals.

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klmGCE Further Mathematics (6370)
5.1 Introduction and revision

Further Pure 2 (MFP2) Textbook

You should have already met the inverse trigonometrical functions when you were studying the A2 specification module Core 3. However, in order to present a clear picture, and for the sake of completeness some revision is included in this section. If y = sin x , we write x = sin ?1 y (or arc sin y ) . Note that sin ?1 y is not cosec y which would normally be written as (sin y ) ?1 when expressed in terms of sine. The use of the superscript -1 is merely the convention we use to denote an inverse in the same way as we say that f ?1 is the inverse of the function f . The sketch of y = sin x will be familiar to you and is shown below.

For any given value of x there is only one corresponding value of y , but for any given value of y there are infinitely many values of x . The graph of y = sin ?1 x being the inverse, is the reflection of y = sin x in the line y = x and a sketch of it is as shown.

As it stands, for a given value of x , y = sin ?1 x has infinitely many values, but if we wish to describe sin ?1 x as a function, we must make sure that the function has precisely one value. In order to overcome this obstacle, we restrict the range of y to ? π ≤ y ≤ π so that the 2 2 ?1 sketch of y = sin x becomes the sketch shown.

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klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

By doing this, we ensure that for any given value of x there is a unique value of y for which y = sin ?1 x . This value is usually called the principal value.

sin ?1 x is the angle between ? 1 π and 1 π inclusive whose sine is x. 2 2

Notice that the gradient of y = sin ?1 x is always greater than zero. We can define cos ?1 x in a similar way but with an important difference. The sketches of y = cos x and y = cos ?1 x are shown below.

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klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

In this case, it would not be sensible to restrict y to values between ? π and π since for 2 2 every value of x ≥ 0 there would be two values of y and for values of x < 0 there would be no value of y. Instead we choose the range 0 ≤ x ≤ π and the sketch is as shown.

cos ?1 x is the angle between 0 and π inclusive whose cosine is x.

When it comes to tan ?1 x we can restrict the range to ? π and π . 2 2
tan ?1 x is the angle between ? π and + π inclusive whose tangent is x. 2 2

The sketch of y = tan ?1 x is shown below.

Exercise 5A
1. Express in terms of π the values of: (a) tan ?1 1 (d) cos ?1 0 (b) cos ?1 3 2 ? ? (e) tan ?1 ? ? 1 ? 3? ? (c) sin ?1 ? 1 2

( )

(f) cos ?1 (?1)

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klmGCE Further Mathematics (6370)
5.2
Suppose then and, differentiating implicitly,
cos y dy =1 dx

Further Pure 2 (MFP2) Textbook

The derivatives of standard inverse trigonometrical functions
y = sin ?1 x sin y = x

thus
dy = 1 dx cos y 1 =

1 ? sin 2 y 1 = 1 ? x2

using cos 2 y + sin 2 y = 1

Note that we choose the positive square root. This is due to the fact that the gradient of the graph of y = sin ?1 x is always greater than zero as was shown earlier.

If

y = sin ?1 x dy 1 = dx 1 ? x2

For y = cos ?1 x using similar working we would arrive at

dy = ? 1 , this time choosing dx 1 ? x2 the negative sign of the square root as the graph of y = cos ?1 x always has a gradient less than zero.

If y = cos ?1 x dy =? 1 dx 1 ? x2

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klmGCE Further Mathematics (6370)
If then we write and, differentiating implicitly,
sec2 y

Further Pure 2 (MFP2) Textbook

y = tan ?1 x tan y = x ,

or

dy =1 dx dy = 12 dx sec y 1 = 1 + tan 2 y using sec 2 y = 1 + tan 2 y dy = 1 dx 1 + x 2

If y = tan ?1 x dy = 1 dx 1 + x 2

Exercise 5B
1. Prove that if y = cos ?1 x then
dy =? 1 . dx 1 ? x2

90

klmGCE Further Mathematics (6370)
5.3

Further Pure 2 (MFP2) Textbook

Applications to more complex differentiation

Some methods of differentiation you should already be familiar with. These would include the function of a function rule, and the product and quotient rules. We will complete this section by using the rules with functions involving inverse trigonometrical functions.

Example 5.3.1
If y = sin ?1 (2 x ? 1) , find
dy . dx

Solution
Set u = 2 x ? 1 then y = sin ?1 u
1 du = 2 and dy = dx du 1? u2

Thus,

dy dy du 2 = × = dx d u d x 1? u2 2 =

1 ? (2 x ? 1) 2 (using the function of a function rule). 2 = 4 x ? 4 x2 2 = 2 x ? x2 1 = x ? x2

Example 5.3.2
Differentiate sin ?1 e x . Set u = e x and let y = sin ?1 e x so that y = sin ?1 u dy 1 du = e x = dx du 1? u2 and using the function of a function rule,
dy dy du = × = ex × 1 dx du d x 1? u2 x = e

1 ? e2 x

91

klmGCE Further Mathematics (6370)
Example 5.3.3
If y = x 2 tan ?1 2 x , find
dy . dx

Further Pure 2 (MFP2) Textbook

This time we need to use the product rule and the function of a function rule. y = x 2 tan ?1 2 x
dy = 2 x tan ?1 2 x + x 2 d tan ?1 2 x dx dx

(

)

For d tan ?1 2 x , set u = 2 x then d tan ?1 2 x = d tan ?1 u × du dx dx du dx = 1 ×2 1+ u2 = 2 2 1 + 4x

(

)

(

)

(

)

Thus

2 dy = 2 x tan ?1 2 x + 2 x 2 . dx 1 + 4x

Example 5.3.4
?1 Differentiate cos x 1 ? x2 ?1 If y = cos x 1 ? x2

Then, using the quotient rule,
? 1 ? x2 ? ? 1 ? 2 dy ? 1? x = dx ? ? cos ?1 x × 1 1 ? x 2 ? ? 2 ? 1 ? x2

(

)

?1 2

× ( ?2 x )

(

)

?1 . = ?1 2 + x cos x 3 1? x 2 2 1? x

(

)

92

klmGCE Further Mathematics (6370)
Exercise 5C
Differentiate the following: 1. (a) tan ?1 3 x 2. (a) x tan ?1 x
?1 3. (a) sin 3 3 x x

Further Pure 2 (MFP2) Textbook

(b) cos ?1 ( 3 x ? 1) (b) e x cos ?1 2 x tan ?1 3 x 2 + 1 1+ x
2

(c) sin ?1 2 x (c) x 2 sin ?1 ( 2 x ? 3)

(b)

(

)

4. (a) sin ?1 ( ax + b )

(b) tan ?1 ( ax + b ) where a and b are positive numbers.

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klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

5.4 Standard integrals integrating to inverse trigonometrical functions
Generally speaking, as you have been taught, formulae from the Formulae and Statistical Tables Booklet supplied for each AS and A2 module apart from MPC1 can be quoted without proof. However, this does not preclude a question requiring a proof of a result from this booklet being set. There are two standard results, the proofs of which are given here and the methods for these proofs should be committed to memory. The first one is

∫ a 2 + x2
Let x = a tan θ so that dx = a sec2 θ dθ Then dx = a sec 2 θ dθ ∫ a 2 + x 2 ∫ a 2 + a 2 tan 2 θ 2 θ dθ = ∫ a sec 2 a sec 2 θ = ∫ 1 dθ a 1 = θ +c a 1 = tan ?1 x + c a a

dx

This integral requires a substitution.

∫ a 2 + x 2 = a tan
The second integral is

dx

1

?1

x +c a



dx a ? x2
2

This interval also requires a substitution Let x = a sin θ Then



dx a ?x
2 2

=

dx = a cos θ dθ a cos θ dθ



= ∫ a cos θ d θ a cos θ =θ +c = sin ?1 x + c a

a 2 ? a 2 sin 2 θ

()



dx a ?x
2 2

= sin ?1 x + c a

()

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klmGCE Further Mathematics (6370)
We now give two examples of definite integrals.

Further Pure 2 (MFP2) Textbook

Example 5.4.1
Evaluate

∫0

2

dx 4 + x2 We have dx ? 1 tan ?1 x ? ∫ 0 4 + x2 = ? ?2 ?0 2? = 1 tan ?1 1 ? 1 tan ?1 0 2 2 1 π = × ?0 2 4 =π 8
2 2

Example 5.4.2
Evaluate



3 2 0

dx 9 ? x2 We have



3 2 0

= 9 ? x2 ?3? ?1 ? 2 ? = sin ? ? ? sin ?1 0 ?3? ? ? = sin ?1 1 ? sin ?1 0 2 π = ?0 6 =π 6

dx

?sin ?1 x ? 2 ? 3 ? ? ?0

()

3

Exercise 5D
Integrate the following, leaving your answers in terms of π . 1.

∫1

3

2dx 1 + x2

2. dx

∫1 2

1

3dx 1? x
2

3. dx x +1
2

∫ ?3

4

dx 25 ? x 2

4.

∫ 0 1 + x2

1

5.

∫? 1

3

3

95

klmGCE Further Mathematics (6370)
5.5 Applications to more complex integrals

Further Pure 2 (MFP2) Textbook

In this section we will show you by means of examples how unfamiliar integrals can often be reduced to one or perhaps two, standard integrals. Most will involve completing the square of a quadratic expression, a method you will no doubt have used many times before in other contexts. We will begin by using examples of integrals which in whole or part reduce to dx ∫ a2 + x2 .

Example 5.5.1
Find

∫ x2 + 4 x + 8 .
2

dx

Now x 2 + 4 x + 8 = ( x + 2 ) + 4 on completing the square so that

∫ x 2 + 4 x + 8 = ∫ ( x + 2 )2 + 4
The substitution u = x + 2 gives du = dx and becomes

dx

dx

( x + 2) + c . therefore 1 tan ?1 u + c or expressing it in terms of x , 1 tan ?1 2 2 2 2
Example 5.5.2
Find

∫ u2 + 4

du

a standard form. The result is

∫ 4 x2 + 4 x + 2
2

dx

We write 4 x 2 + 4 x + 2 = ( 2 x + 1) + 1 so that we have

∫ ( 2 x + 1)2 + 1 .

dx

The substitution

u = 2 x + 1 gives du = 2dx and the integral becomes 1 du ∫ u22 + 1 = 1 tan ?1 u + c 2 = 1 tan ?1 ( 2 x + 1) + c or substituting back 2

96

klmGCE Further Mathematics (6370)
Example 5.5.3
Find

Further Pure 2 (MFP2) Textbook

∫ x4 + 9

xdx

Here the substitution u = x 2 transforms the given integral into standard form for if u = x 2 1 du du = 2 xdx and we have ∫ 2 u2 + 9 = 1 × 1 tan ?1 u + c 2 3 3 2 = 1 tan ?1 x + c 6 3

Finally we give a slightly harder example of an integral which uses

∫ a2 + x2

dx

in its solution.

Example 5.5.4
Find

∫ x 2 + 6 x + 12 dx .

x+5

Integrals of this type where the numerator is a linear expression in x and the denominator is a quadratic in x usually integrate to ln p ( x) + tan ?1 q ( x) where p ( x) and q ( x) are functions of
f ' ( x ) dx x. In order to tackle this integral you need to remember that integrals of the form ∫ f ( x)

integrate to ln f ( x ) + c . You should have been taught this result when studying the module Core 3. So to start evaluating this integral we have to note that the derivative of x 2 + 6 x + 12 is 2 x + 6 and we rewrite the numerator of the integral as 1 ( 2 x + 6 ) + 2 so that the integral becomes 2



1 2x + 6 + 2 ( ) 2 dx x 2 + 6 x + 12

and separating it into two halves we write it again as



1 2 x + 6 dx ( ) 2 + ∫ 2 2dx 2 x + 6 x + 12 x + 6 x + 12

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klmGCE Further Mathematics (6370)
( )

Further Pure 2 (MFP2) Textbook

The first integral integrates to 1 ln x 2 + 6 x + 12 whilst in the second we complete the square 2 in the denominator and write it as

∫ ( x + 3)2 + 3
The substitution u = x + 3 leads to

2dx

∫ u2 + 3
= 2 tan ?1 u 3 3 ? ? = 2 tan ?1 ? x + 3 ? 3 ? 3 ? So that

2du

∫ x 2 + 6 x + 12 dx ∫

x+5

=

1 ln x 2 + 6 x + 12 + 2 tan ?1 ? x + 3 ? + c ? ? 2 3 ? 3 ?

(

)

The final part of this section will show you how integrals can sometimes be reduced to dx . This will be done by means of examples. a 2 ? x2

Example 5.5.5
Find



dx 4x ? x
2

.

As in previous examples, we need to complete the square on 4 x ? x 2 , and we write
4 x ? x2 = 4 ? ( x ? 2)
2

So that the interval becomes



dx 4 ? ( x ? 2)
2

98

klmGCE Further Mathematics (6370)

du 4 ? u2

Further Pure 2 (MFP2) Textbook

The substitution u = x ? 2 simplifies the result to the standard form

= sin ?1 u + c 2 = sin ?1

( x ? 2) + c
2

Example 5.5.6
Find



dx 1 + 6 x ? 3x2

.

In order to complete the square in the denominator, we write 6 x ? 3x 2 = 3 2 x ? x 2 = 1 + 3 1 ? ( x ? 1)
= 4 ? 3 ( x ? 1)
2

(

(

2

)

)

Thus,



dx 1 + 6 x ? 3x
2

=



dx 4 ? 3 ( x ? 1)
2

The substitution of u = 3 ( x ? 1) reduces the integral to

3 4 ? u2 = 1 sin ?1 u + c 2 3 3 ( x ? 1) = 1 sin ?1 +c 2 3 One final example shows how more complicated expressions may be integrated using methods shown here and other results which you should have met studying earlier modules. In this particular context the result you will need is that



du

∫ f ( x ) dx
and the integral becomes

f ' ( x)

= 2 f ( x) + c

[This result can be easily verified using the substitution u = f ( x ) , since then, du = f ' ( x ) dx



du ]. u 99

klmGCE Further Mathematics (6370)
Example 5.5.7
Find

Further Pure 2 (MFP2) Textbook



xdx 7 ? 6x ? x
2

.

Now the derivative of 7 ? 6 x ? x 2 is ?6 ? 2 x so we write x as ? 1 ( 6 + 2 x ) + 3 and the integral 2 becomes ? 1 (6 + 2x) + 3 dx ?∫ 2 7 ? 6 x ? x2 or separating the integral into two parts ? 1 (6 + 2x) ?∫ 2 7 ? 6 x ? x2 The first integral is of the form f ' ( x)

?



3dx 7 ? 6 x ? x2

∫ f ( x ) dx apart from a scaler multiplier, and so integrates to

7 ? 6 x ? x 2 , whilst completing the square on the denominator of the second integral, we 3dx which integrates to 3sin ?1 x + 3 using the substitution u = x + 3 . obtain ∫ 2 4 16 ? ( x + 3) Hence,



xdx 7 ? 6x ? x
2

= ? 7 ? 6 x ? x 2 ? 3sin ?1 x + 3 + c 4

( )

100

klmGCE Further Mathematics (6370)
Exercise 5E
1. Integrate (a) 1 x + 4x + 5
2

Further Pure 2 (MFP2) Textbook

(b)

1 2x ? 4x + 5
2

(c)

1 x ?x+2
2

2. Integrate (a) 3. Find (a) 2x x + 2x + 3
2

(b)

x x + x +1
2



dx 7 + 6x ? x
2

(b)



dx 3 + 2x ? x
2

(c)



dx x (1 ? 2 x )

4. Find (a)



x + 1 dx 1 ? x2

(b)



3x ? 2 3 + 2 x ? x2

dx

(c)



(1 ? x )
1 ? x ? x2

dx

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klmGCE Further Mathematics (6370)
Chapter 6: Hyperbolic Functions
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 Definitions of hyperbolic functions Numerical values of hyperbolic functions Graphs of hyperbolic functions Hyperbolic identities Osborne’s rule Differentiation of hyperbolic functions Integration of hyperbolic functions Inverse hyperbolic functions Logarithmic form of inverse hyperbolic functions

Further Pure 2 (MFP2) Textbook

6.10 Derivatives of inverse hyperbolic functions 6.11 Integrals which integrate to inverse hyperbolic functions 6.12 Solving equations

This chapter introduces you to a wholly new concept. When you have completed it, you will: ? ? ? ? ? ? ? ? know what hyperbolic functions are; be able to sketch them; be able to differentiate and integrate them; have learned some hyperbolic identities; understand what inverse hyperbolic functions are and how they can be expressed in alternative forms; be able to sketch inverse hyperbolic functions; be able to differentiate inverse hyperbolic functions and recognise integrals which integrate to them; be able to solve equations involving hyperbolic functions.

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klmGCE Further Mathematics (6370)
6.1 Definitions of hyperbolic functions

Further Pure 2 (MFP2) Textbook

It was shown in Chapter 4 that sin x = 1 eix ? e ?ix and cos x = 1 eix +e ?ix . Hyperbolic 2i 2 functions are defined in a very similar way. The definitions of sinh x and cosh x (often called hyperbolic sine and hyperbolic cosine – pronounced ‘shine x’ and ‘cosh x’) are:
sinh x = 1 e x ? e ? x 2 1 cosh x = e x + e? x 2

(

)

(

)

( (

) )

There are four other hyperbolic functions derived from these, just as there are four trigonometric functions. They are:

tanh x = sinh x cosh x cosech x = 1 sinh x sech x = 1 cosh x coth x = 1 tanh x
In terms of exponential functions,

‘than x’ ‘cosheck x’ ‘sheck x’ ‘coth x’

tanh x = sinh x cosh x =
1 2 1 2

(e (e

x x

? e? x + e? x

) )

x ?x = e x ? e? x , e +e or, on multiplying the numerator and denominator by e x , 2x tanh x = e2 x ? 1 . e +1

Again,

cosech x =

1 sinh x 1 = x ?x 1 2 e ?e

(

)

=

2 . e ? e? x
x

Exponential forms for sech x and coth x can be found in a similar way.

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klmGCE Further Mathematics (6370)
Exercise 6A
1. Express, in terms of exponentials: (a) sech x, (b) coth x, x, (c) tanh 1 2

Further Pure 2 (MFP2) Textbook

(d) cosech 3x.

6.2

Numerical values of hyperbolic functions

When finding the value of trigonometric functions, for example sin x, the angle x must be given in degrees (or radians). There is no unit for x when evaluating, for example, sinh x. It is quite in order to speak about sinh 2 or cosh1.3 :
2 ?2 sinh 2 = e ? e = 3.63 (to two decimal places); 2 ?1.3 1.3 cosh1.3 = e + e = 1.97 (to two decimal places). 2 You can work out these values on a calculator using the e x button. However, for convenience most scientific calculators have a ‘hyp’ button and sinh 2 can be obtained directly by pressing the ‘2’, ‘hyp’ and ‘sin’ buttons in the appropriate order.

It is worth remembering that 0 ?0 sinh 0 = e ? e = 1 ? 1 = 0; 2 2 ?0 0 cosh 0 = e + e = 1 + 1 = 1. 2 2

Exercise 6B
1. Use a calculator to evaluate, to two decimal places: (a) sinh 0.6, (d) tanh (– 0.6), (b) tanh 1.3, (e) cosh (– 0.3), (c) sech 2.1, (f) coth 4

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klmGCE Further Mathematics (6370)
6.3 Graphs of hyperbolic functions

Further Pure 2 (MFP2) Textbook

The graphs of hyperbolic functions can be sketched easily by plotting points. Some sketches are given below but it would be a good exercise to make a table of values and confirm the general shapes for yourself. It would also be worthwhile committing the general shapes of y = sinh x, y = cosh x and y = tanh x to memory.
y

y

y = sinh x
0

x

y = cosh x
1 0
x

The sketch of y = tanh x requires a little more consideration. In Section 5.1, it was shown that tanh x could be written as 2x tanh x = e2 x ? 1 e +1 2x ? ? = ? ? 1 ? e2 x ? . ? 1+ e ? Now, as e 2 x > 0 for all values of x, it follows that the numerator in the bracketed expression above is less than its denominator, so that tanh x > ?1. Also, as x → ?∞, e 2 x → 0 and tanh x → ? 1 = ?1. So the graph of y = tanh x has an asymptote at y = ?1. 1

()

Now, if the numerator and denominator of tanh x are divided by e 2 x , tanh x = 1 ? e ?2 x . 1+ e ?2 x As e > 0 for all values of x, it follows that the numerator of this fraction is less than its denominator, from which it can be deduced that tanh x < 1. It can also be deduced that as e ?2 x → 0 as x → ∞, so tanh x → 1 as x → ∞. So the graph of y = tanh x has an asymptote at x = 1. Hence the curve y = tanh x lies between y = +1 and y = ?1, and has y = ±1 as asymptotes.
y = tanh x
y
?2 x

1 0 –1
x

105

klmGCE Further Mathematics (6370)
Exercise 6C
1. Sketch the graphs of (a) y = sech x, (b) y = cosech x, (c) y = coth x.

Further Pure 2 (MFP2) Textbook

6.4

Hyperbolic identities

Just as there are trigonometric identities such as cos 2 θ + sin 2 θ = 1 and cos 2θ = 2 cos 2 θ ? 1, there are similar hyperbolic identities. For example, ? x ?x ? cosh x = ? e + e ? = 1 e2 x + 2 + e?2 x , 2 ? 4 ?
2 2

(

)

? x ?x ? and sinh x = ? e ? e ? = 1 e2 x ? 2 + e ?2 x 2 ? 4 ? from which, by subtraction,
2

2

(

) )

cosh 2 x ? sinh 2 x = 1 e 2 x + 2 + e ?2 x ? 1 e2 x ? 2 + e ?2 x 4 4 = 1.
cosh 2 x ? sinh 2 x = 1

(

) (

Dividing both sides of this equation by cosh 2 x, it follows that cosh 2 x ? sinh 2 x = 1 2 2 cosh x cosh x cosh 2 x 1 ? tanh 2 x = sech 2 x. sech 2 x = 1 ? tanh 2 x Or again, dividing both sides by sinh 2 x instead, cosh 2 x ? sinh 2 x = 1 sinh 2 x sinh 2 x sinh 2 x coth 2 x ? 1 = cosech 2 x. cosech 2 x = coth 2 x ? 1

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klmGCE Further Mathematics (6370)
(

Further Pure 2 (MFP2) Textbook

Now consider sinh x cosh y + cosh x sinh y = 1 e x ? e? x 1 e y ? e? y + 1 e x + e? x 1 e y ? e? y 2 2 2 2 1 x ?y ?x ? y x y x y ?x y = e e ? e e + e e ? e e + e e + e? x e y ? e x e? y ? e? x e? y 4 = 1 2e x e y ? 2e ? x e ? y 4 = 1 e x e y ? e? x e? y 2 ? x+ y = 1 ex+ y ? e ( ) [using the laws of indices] 2 = sinh ( x + y ) .

(

) (

) (

) (

)

)

( (

)

)

(

)

In exactly the same way, expressions for sinh ( x ? y ) , cosh ( x + y ) and cosh ( x ? y ) can be worked out. sinh ( x ± y ) = sinh x cosh y ± cosh x sinh y cosh ( x ± y ) = cosh x cosh y ± sinh x sinh y

Exercise 6D
1. Show that (a) sinh ( x ? y ) = sinh x cosh y ? cosh x sinh y, (b) cosh ( x ± y ) = cosh x cosh y ± sinh x sinh y. You will probably remember that the basic trigonometric formulae for sin ( x + y ) and
cos ( x + y ) can be used to find expressions for sin 2 x, cos 2 x, and so on. The hyperbolic

formulae given above help to find corresponding results for hyperbolic functions.

107

klmGCE Further Mathematics (6370)
For example, because sinh ( x + y ) = sinh x cosh y + cosh x sinh y, putting y = x,
sinh ( x + x ) = sinh x cosh x + cosh x sinh x,

Further Pure 2 (MFP2) Textbook

or Using putting y = x,

sinh 2 x = 2sinh x cosh x. cosh ( x + y ) = cosh x cosh y + sinh x sinh y, cosh ( x + x ) = cosh x cosh x + sinh x sinh x,

or

cosh 2 x = cosh 2 x + sinh 2 x.

Using cosh 2 x ? sinh 2 x = 1, cosh 2 x = 1 + sinh 2 x + sinh 2 x or = 1 + 2sinh 2 x cosh 2 x = 2 cosh 2 x ? 1.
sinh 2 x = 2sinh x cosh x cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 x ? 1 = 1 + 2sinh 2 x

Some examples will illustrate extensions of these results.

108

klmGCE Further Mathematics (6370)
Example 6.4.1
x . Show that tanh 2 x = 2 tanh 2 1 + tanh x

Further Pure 2 (MFP2) Textbook

Solution
tanh 2 x = sinh 2 x cosh 2 x x cosh x . = 2sinh 2 cosh x + sinh 2 x Dividing the numerator and denominator by cosh 2 x, 2sinh x cosh x cosh 2 x tanh 2 x = cosh 2 x + sinh 2 x cosh 2 x 2sinh x = cosh x 2 1 + sinh 2 x cosh x 2 tanh x . = 2 1 + tanh x

Example 6.4.2
Show that cosh 3x = 4 cosh 3 x ? 3cosh x.

Solution
cosh 3x = cosh ( 2 x + x ) = cosh 2 x cosh x + sinh 2 x sinh x = 2 cosh 2 x ? 1 cosh x + 2sinh x cosh x sinh x = 2 cosh 3 x ? cosh x + 2 cosh 2 x ? 1 cosh x = 2 cosh 3 x ? cosh x + 2 cosh 3 x ? 2 cosh x = 4 cosh 3 x ? 3cosh x.

(

)

(

)

[using cosh x ? sinh x = 1]

2

2

Exercise 6E
1. Using the expansion of sinh ( 2 x + x ) , show that sinh 3 x = 3sinh x + 4sinh 3 x. 2. Express cosh 4 x in terms of cosh x.

109

klmGCE Further Mathematics (6370)
6.5 Osborne’s rule

Further Pure 2 (MFP2) Textbook

It should be clear that the results and identities for hyperbolic functions bear a remarkable similarity to the corresponding ones for trigonometric functions. In fact the only differences are those of sign – for example, whereas cos 2 x + sin 2 x = 1, the corresponding hyperbolic identity is cosh 2 x ? sinh 2 x = 1. There is a rule for obtaining the identities of hyperbolic functions from those for trigonometric functions – it is called Osborne’s rule. To change a trigonometric function into its corresponding hyperbolic function, where a product of two sines appears change the sign of the corresponding hyperbolic term For example, because then Note also that because then

cos ( x + y ) = cos x cos y ? sin x sin y cosh ( x + y ) = cosh x cosh y + sinh x sinh y.

cos 2 x = 1 ? 2sin 2 x cosh 2 x = 1 + 2sinh 2 x,

because sin 2 x is a product of two sines. However, care must be exercised in using this rule, as the next example shows. It is known that sec 2 x = 1 + tan 2 x but sech 2 x = 1 ? tanh 2 x The reason that the sign has to be changed here is that a product of sines is implied because 2 tan 2 x = sin 2 x . cos x It should be noted that Osborne’s rule is only an aid to memory. It must not be used in a proof – for example, that cosh 2 x ? sinh 2 x = 1. The method shown in Section 5.4 must be used for that.

110

klmGCE Further Mathematics (6370)
6.6 Differentiation of hyperbolic functions

Further Pure 2 (MFP2) Textbook

You will already have met the derivative of ekx . Just to remind you, d e kx = ke kx . dx

( )

As hyperbolic functions can be expressed in terms of e, it follows that their differentiation is straightforward. For example,
y = sinh x = 1 e x ? e? x . 2

(

)

Therefore,

dy 1 x ? x = e +e dx 2 = cosh x.

(

) )

And if

y = sinh kx = 1 e kx ? e ? kx , 2

(

then

dy 1 = ke kx +ke ? kx dx 2 = k 1 e kx +e ? kx 2 = k cosh kx.

(

(

) )

111

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

You can differentiate cosh x and cosh kx in exactly the same way; tanh x can be differentiated by treating it as the derivative of the quotient sinh x . The following results cosh x should be committed to memory: y = sinh x, y = cosh x, y = tanh x, dy = cosh x dx dy = sinh x dx dy = sech 2 x dx

Generally: y = sinh kx, y = cosh kx, y = tanh kx, dy = k cosh kx dx dy = k sinh kx dx dy = k sech 2 kx dx

Note that the derivatives are very similar to the derivatives of trigonometric functions, except that whereas d ( cos x ) = ? sin x, d ( cosh x ) = + sinh x. dx dx

Example 6.6.1
Differentiate sinh 1 x. 2

Solution
y = sinh 1 x 2 dy = cosh 1 x × 1 dx 2 2 = 1 cosh 1 x. 2 2

( )

112

klmGCE Further Mathematics (6370)
Example 6.6.2
Differentiate x cosh 2 x + cosh 4 3 x.

Further Pure 2 (MFP2) Textbook

Solution
y = x cosh 2 x + cosh 4 3 x = x cosh 2 x + ( cosh 3x ) .
4

Using the product and chain rules for differentiation, dy 3 = (1× cosh 2 x ) + ( x sinh 2 x × 2 ) + 4 ( cosh 3x ) × sinh 3 x × 3 dx = cosh 2 x + 2 x sinh 2 x + 12sinh 3 x cosh 3 3 x.

(

)

Exercise 6F
1. Differentiate the following expressions: (a) cosh 3 x, (d) cosh 2 x , x (g) cosech x. (b) cosh 2 3x, (e) x tanh x, (c) x 2 cosh x, (f) sech x [hint: write sech x as ( cosh x ) ] ,
?1

113

klmGCE Further Mathematics (6370)
6.7 Integration of hyperbolic functions
1 kx kx ∫ e dx = k e .

Further Pure 2 (MFP2) Textbook

You will already have met the integral of e kx . Just to remind you,

As hyperbolic functions can be expressed in terms of e, and as it is the reverse of differentiation, it follows that their integration is straightforward.

∫ sinh x dx = cosh x + c ∫ cosh x dx = sinh x + c
2 ∫ sech x dx = tanh x + c

Of course, generally ∫ sinh kx = 1 cosh kx + c. k For the integration of tanh x, a substitution is needed as follows: tanh x dx = sinh x dx. cosh x





Putting u = cosh x, So that

du = sinh x dx.
du

∫ tanh x dx = ∫ u

= ln u + c = ln cosh x + c.

∫ tanh x dx = ln cosh x + c ∫ coth x dx = ln sinh x + c
Exercise 6G
1. Evaluate the following integrals: (a) (c)

∫ cosh 3x dx, ∫ x sinh 2 x dx,

(b) (d)

∫ cosh ∫ tanh

2

x dx, x dx.

2

114

klmGCE Further Mathematics (6370)
6.8 Inverse hyperbolic functions

Further Pure 2 (MFP2) Textbook

Just as there are inverse trigonometric functions sin ?1 x, cos ?1 x, etc. , so there are inverse hyperbolic functions. They are defined in a similar way to inverse trigonometric functions – so, if x = sinh y, then y = sinh ?1 x; and likewise for the other five hyperbolic functions. To find the value of, say, sinh ?1 2 using a calculator, you use it in the same way as you would if it was a trigonometric function (pressing the appropriate buttons for hyperbolic functions). The sketches of y = sinh ?1 x and y = tanh ?1 x are the reflections of y = sinh x and y = tanh x, respectively, in the line y = x. These sketches are shown below. Note that the curve y = tanh ?1 x has asymptotes at x = ±1.
y = tanh
y = sinh
?1

(

)

y
?1

x

y

x

0
x

–1

0

1

x

Note also that y = cosh x does not have an inverse. This is because the mapping f : x → cosh x is not a one-to-one mapping. If you look at the graph of y = cosh x (in Section 5.3) you will see that for every value of y > 1 there are two values of x. However, if the domain of y = cosh x is restricted to x ≥ 0 there will be a one-to-one mapping, and hence an inverse, and the range for the inverse will be y ≥ 0.

115

klmGCE Further Mathematics (6370)
6.9

Further Pure 2 (MFP2) Textbook

Logarithmic form of inverse hyperbolic functions

The inverse hyperbolic functions cosh ?1 x, sinh ?1 x and tanh ?1 x can be expressed as logarithms. For example, if y = cosh ?1 x, then x = cosh y
y ?y x = e +e 2 y 2 x = e + e? y . 2 xe y = e 2 y + 1

Multiplying by e y ,

0 = e 2 y ? 2 xe y + 1. This is a quadratic equation in e y and can be solved using the quadratic formula:
2 e y = 2x ± 4x ? 4 2

= x ± x 2 ? 1.

Taking the logarithm of each side,

y = ln x ± x 2 ? 1 . Now,

(

)

(

x + x2 ? 1 x ? x2 ? 1 = x2 ?

)(

)

(
)

x2 ? 1

)

2

= x2 ? x2 + 1 = 1. Thus x ? x2 ? 1 =

(x +

1 x ?1
2

, ? ? ? ? ?

and

? ? 1 ln x ? x ? 1 = ln ? 2 ? x + x ?1 ?

(

2

so that

( = ? ln ( x + y = ± ln ( x +

)

) x ?1) , x ?1).
2 2

116

klmGCE Further Mathematics (6370)
However, from Section 5.8, if y ≥ 0 then y = cosh ?1 x = + ln x + x 2 ? 1 .

Further Pure 2 (MFP2) Textbook

(

)

A similar result for y = sinh ?1 x can be obtained by writing x = sinh y and then expressing sinh y in terms of e y . This gives y = ln x ± x 2 + 1 , but as x ? x 2 + 1 ≤ 0 the negative sign has to be rejected because the logarithm of a negative number is non-real. Thus sinh ?1 x = ln x + x 2 + 1 . It is straightforward to obtain the logarithmic form of y = tanh ?1 x if, after writing sinh y tanh y = x, tanh y is written as . cosh y cosh ?1 x = ln x + x 2 ? 1 sinh ?1
2

(

)

(

)

( x = ln ( x +

tanh ?1 x = 1 ln 1 + x 2 1? x

( )

) x + 1)

117

klmGCE Further Mathematics (6370)
Example 6.9.1
Find, in logarithmic form, tanh ?1 1 . 2

Further Pure 2 (MFP2) Textbook

Solution
? 1+ 1 ? 2 tanh ?1 1 = 1 ln ? ? 1 ? ? 2 2 ? 1? 2 ? ?3 ? 2 = 1 ln ? ? 1 ? 2 ? ?2? = 1 ln 3 2 = ln 3.

Example 6.9.2
Find, in logarithmic form, sinh ?1 3 . 4

Solution
2 ? ? sinh ?1 3 = ln ? 3 + 3 + 1 ? 4 4 ?4 ? ? ? ? ? = ln ? 3 + 9 + 1 ? 16 ? ?4

()

? ? = ln ? 3 + 25 ? 4 16 ? ? = ln 3 + 5 4 4 = ln 2.

( )

Exercise 6H
1. Show that sinh ?1 x = ln x + x 2 + 1 . 2. Show that tanh ?1 x = 1 ln 1 + x . 2 1? x 3. Express the following in logarithmic form: (a) cosh ?1 1.5, (b) tanh ?1 1 , (c) sinh ?1 5 . 3 12

(

)

( )

118

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

6.10 Derivatives of inverse hyperbolic functions
As already seen, if y = sinh ?1 then sinh y = x. Differentiating with respect to x, dy = 1. cosh y dx dy So that = 1 dx cosh y 1 ?using cosh 2 y ? sinh 2 y = 1? = ? ? 2 1+ x Again, if y = sinh ?1 x , a sinh y = x , a and Thus
cosh y dy 1 = . dx a

dy 1 = dx a cosh y 1 =

2 a 1 + x2 a 1 = . 2 a + x2

The derivatives of cosh ?1 x and cosh ?1 x , and also tanh ?1 x and tanh ?1 x are obtained a a in exactly the same way.

()

()

Example 6.10.1
Find the derivative of d tanh ?1 x . dx

(

)

Solution
y = tanh ?1 x

tanh y = x sech 2 y dy =1 dx dy = 12 dx sech y = 1 2 1? x

?using sech 2 y + tanh 2 y = 1? . ? ?

119

klmGCE Further Mathematics (6370)
y = sinh ?1 x : y = cosh ?1 x : y = tanh ?1 x :

Further Pure 2 (MFP2) Textbook

It is suggested that you work through the remaining results yourself.
dy 1 = dx 1 + x2 dy 1 = dx x2 ? 1 dy = 1 dx 1 ? x 2

Generally,
y = sinh ?1 x : a y = cosh ?1 x : a y = tanh ?1 x : a

dy 1 = 2 dx a + x2 dy 1 = 2 dx x ? a2 dy = 1 dx a 2 ? x 2

Example 6.10.2
Differentiate cosh ?1 x . 3

Solution
y = cosh ?1 x 3 dy 1 = 2 dx x ? 32 =

1
x2 ? 9

.

120

klmGCE Further Mathematics (6370)
Example 6.10.3

Further Pure 2 (MFP2) Textbook

dy If y = x 2 sinh ?1 x , find when x = 2, giving your answer in the form a + b ln c. 2 dx

Solution
Differentiate x 2 sinh ?1 x as a product . 2 y = x 2 sinh ?1 x 2 dy 1 = 2 x sinh ?1 x + x 2 . 2 dx 2 2 + x2 So, when x = 2 dy 1 = 4sinh ?1 1 + 4 2 dx 2 + 22
= 4 ln 1 + 1 + 1 + 4 8 = 4 ln 1 + 2 + 4 8 8× 8

( (

)

)

( ) = 4 ln (1 + 2 ) + = 4 ln (1 + 2 ) +
Exercise 6I

= 4 ln 1 + 2 + 4 8 8 8 2 2.

1. Differentiate the following: (a) tanh ?1 x , (b) sinh ?1 x , 3 3 (e) 1 cosh ?1 x 2 . (d) e x sinh ?1 x, x 2. If y = x cosh ?1 x, find

(c) cosh ?1 x , 4

dy when x = 2, giving your answer in the form a + ln b, where a dx and b are irrational numbers.

121

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

6.11 Integrals which integrate to inverse hyperbolic functions
Integration can be regarded as the reverse of differentiation so it follows, from the results in Section 6.10, that:

∫ ∫ ∫
The integrals of 1
a +x
2 2

= sinh ?1 x + c a a +x dx = cosh ?1 x + c a x2 ? a2
2 2

dx

= 1 tanh ?1 x + c a a a ?x
2 2

dx

and

1
x ? a2
2

, in particular, help to widen the ability to integrate. 1
px + qx + r
2

In fact, these results can be used to integrate any expression of the form even
sx + t px + qx + r
2

, or

, where p > 0. The examples which follow show how this can be done.

Example 6.11.1
Find



dx 16 + x
2

.

Solution



dx 16 + x
2

=

4 + x2 = sinh ?1 x + c. 4



dx
2

()

122

klmGCE Further Mathematics (6370)
Example 6.11.2
Find

Further Pure 2 (MFP2) Textbook



dx 6 + 2 x2

.

Solution



dx 6 + 2x
2

=



2 3 + x2

(

dx dx

)

= 1 2



3 + x2

? ? = 1 sinh ?1 ? x ? + c. 2 ? 3?

Example 6.11.3
Find



dx x + 2x ? 3
2

.

Solution
In order to evaluate this integral, you must complete the square in the denominator. x2 + 2 x ? 3 = x2 + 2 x + 1 ? 4 = ( x + 1) ? 4.
2

(

)

Hence,

∫ ∫

dx x2 + 2 x ? 3

=

∫ ∫

dx

( x + 1) ? 4
2

.

Substituting z = x + 1, for which dz = dx, gives dx = 2dz . 2 z ?4 x + 2x ? 3

= cosh ?1 z + c 2 = cosh ?1 x + 1 + c. 2

() ( )

123

klmGCE Further Mathematics (6370)
Example 6.11.4
Find

Further Pure 2 (MFP2) Textbook



2x + 5 x + 6 x + 10
2

dx.

Solution
In this case, the integral is split into two by writing one integral with a numerator which is the derivative of x 2 + 6 x + 10. d x 2 + 6 x + 10 = 2 x + 6. Now, dx

(

)

But and

2x + 5 = 2x + 6 ?1



2x + 5 x + 6 x + 10
2

dx =

x + 6 x + 10 = I1 ? I 2 , say.



2x + 6
2

dx ?



1 x + 6 x + 10
2

dx.

Because the derivative of x 2 + 6 x + 10 is 2 x + 6, then for I1 the substitution z = x 2 + 6 x + 10 gives dz = 2 x + 6 and consequently dx I1 = dz z

∫ ∫

= z 2 dz
2 = z1 1

?1

2

= 2 x 2 + 6 x + 10. For I 2 , completing the square in the denominator,
x 2 + 6 x + 10 = x 2 + 6 x + 9 + 1 = ( x + 3) + 1.
2

So that

I2 =

∫ ( x + 3)2 + 1.

dx

The substitution u = x + 3 will give du = 1, dx d u I = and
2



u2 +1

= sinh ?1 u = sinh ?1 ( x + 3) . Therefore, the complete integral is I1 ? I 2 = 2 x 2 + 6 x + 10 ? sinh ?1 ( x + 3) + c.

124

klmGCE Further Mathematics (6370)
Exercise 6J
1. Evaluate the following integrals: dx , (a) (b)

Further Pure 2 (MFP2) Textbook



x +9 dx
2

2



dx x ? 16 dx
2

, ,

(c) (f)

∫ ∫

dx 4 x + 25 dx
2

, ,

(d)

∫ ∫

9 x + 49 dx
2

,

(e)

∫ ∫

( x + 1)
dx
2

2

+4

( x ? 2)

2

? 16

(g)

x + 4x + 5

,

(h)

x ? 2x ? 2

.

6.12 Solving equations
You are likely to meet two types of equations involving hyperbolic functions. The methods for solving them are quite different. The first type has the form a cosh x + b sinh x = c, or a similar linear combination of hyperbolic functions. The correct method for solving this type of equation is to use the definitions of sinh x and cosh x to turn the equation into one involving e x (frequently a quadratic equation).

Example 6.12.1
Solve the equation 7 sinh x ? 5cosh x = ?1.

Solution
x ?x x ?x Using the definitions sinh x = e ? e and cosh x = e + e , 2 2 x ? x x ? x ? ? ? ? 7 ? e ? e ? ? 5 ? e + e ? = ?1 2 ? ? 2 ? ?

7e x ? 7e ? x ? 5e x ? 5e ? x = ?1 2 2 2 2 x e ? 6e ? x = ?1. Multiplying throughout by e x , e 2 x ? 6 = ?e x , or
e 2 x + e x ? 6 = 0.

This is a quadratic equation in e x and factorizes to

(e

x

+ 3 e x ? 2 = 0.

)(

)

Hence, e = ?3 or e = 2. The only solution possible is x = ln 2 because e x ≠ ?3 since e x > 0 for all values of x.

x

x

125

klmGCE Further Mathematics (6370)
Example 6.12.2
Solve the equation cosh 2 x = 4sinh x + 6.

Further Pure 2 (MFP2) Textbook

Solution
The identity cosh 2 x ? sinh 2 x = 1 is used here. The reason for this can be seen on substitution – instead of having an equation involving cosh x and sinh x, the original equation is reduced to one involving sinh x only. 1 + sinh 2 x = 4sinh x + 6 sinh 2 x ? 4sinh x ? 5 = 0. This is a quadratic equation in sinh x which factorizes to ( sinh x ? 5)( sinh x + 1) = 0. Hence, and and
sinh x = 5 or sinh x = ?1 x = sinh ?1 5 or x = sinh ?1 ? 1, x = 2.31 or x = ?0.88 (to two decimal places).

The answers can also be expressed in terms of logarithms, using the results from Section 5.9:
sinh ?1 5 = ln 5 + 52 + 1 = ln 5 + 26 ;
2 ? sinh ?1 ? 1 = ln ? ? ?1 + ( ?1) + 1 ? = ln 2 ? 1 . ? ? Note that it is not advisable to use the definitions of sinh x and cosh x when attempting to solve the equation in Example 5.12.2 – this would generate a quartic equation in e x which would be difficult to solve.

(

) (

)

(

)

Examples 6K
1. Solve the equations: (a) 4sinh x + 3e x = 9, (c) cosh 2 x ? 3sinh x = 5, (e) tanh x = 7sech x + 3.
2

(b) 3sinh x + 4 cosh x = 4, (d) cosh 2 x ? 3cosh x = 4,

126

klmGCE Further Mathematics (6370)
Miscellaneous exercises 6
1. (a) Express cosh x + sinh x in terms of e x . (b) Hence evaluate
[AQA March 1999]

Further Pure 2 (MFP2) Textbook





0

1 dx. cosh x + sinh x

2. (a) Using the definitions prove that

cosh x = 1 e x + e? x and sinh x = 1 e x ? e ? x , 2 2
2sinh x cosh x = sinh 2 x.

(

)

(

)

(b) Hence, or otherwise, solve the equation 8sinh x = 3sech x, leaving your answer in terms of natural logarithms.
[AEB January 1997]

3. (a) By considering sinh y = x. or otherwise, prove that sinh ?1 x = ln x + 1 + x 2 . (b) Solve the equation

(

)

2 cosh 2θ ? 5sinh θ ? 8 = 0,

leaving your answers in terms of natural logarithms.
[AEB January 1996]

4. (a) Show that the equation can be expressed as

14sinh x ? 10 cosh x = 5

2e 2 x ? 5e x ? 12 = 0.

(b) Hence solve the equation 14sinh x ? 10 cosh x = 5, giving your answer as a natural logarithm.
[AQA June 2001]

127

klmGCE Further Mathematics (6370)
4 cosh 3 t ? 3cosh t = cosh 3t.

Further Pure 2 (MFP2) Textbook

5. (a) Starting from the definition of cosh t in terms of et , show that

(b) Hence show that the substitution x = cosh t transforms the equation 16 x3 ? 12 x = 5 cosh 3t = 5 . into 4 (c) The above equation in x has only one real root. Obtain this root, giving your answer in the form 2 p + 2q , where p and q are rational numbers to be found.
[AQA March 1999]

6. (a) Using the definitions of sinh θ and cosh θ in terms of exponentials, show that 2θ tanh θ = e2θ ? 1 . e +1 (b) Hence prove that tanh ?1 x = 1 ln 1 + x , 2 1? x where ?1 < x < 1.

( )

[AEB June 1999]

7. (a) By expressing tanh x in terms of sinh x and cosh x, show that (i) tanh 2 x = 1 ? sech 2 x, (ii) d tanh x = sech 2 x. dx (b) (i) Find

∫ ( tanh x ? tanh x ) dx. (ii) Hence, or otherwise, find tanh ∫
3

3

x dx.

[AQA March 2000]

128

klmGCE Further Mathematics (6370)

Further Pure 2 (MFP2) Textbook

8. (a) Explain, by means of a sketch, why the numbers of distinct values of x satisfying the equation cosh x = k in the cases k < 1, k = 1 and k > 1 are 0, 1 and 2, respectively. (b) Given that cosh x = 17 and sinh y = 4 , 8 3

(i) express y in the form ln n, where n is an integer, (ii) show that one of the possible values of x + y is ln12 and find the other possible value in the form ln a, where a is to be determined.
[AQA March 2000]

9. (a) State the values of x for which cosh ?1 x is defined. (b) A curve C is defined for these values of x by the equation y = x ? cosh ?1 x. (i) Show that C has just one stationary point. (ii) Evaluate y at the stationary point, giving your answer in the form p ? ln q, where p and q are numbers to be determined.
[NEAB March 1998]

10. (a) Prove that d tanh x = sech 2 x. ( ) dx (b) Hence, or otherwise, prove that d tanh ?1 x = 1 . dx 1 ? x2

(

)

(c) By expressing

1 in partial fractions and integrating, show that 1 ? x2 tanh ?1 x = 1 ln 1 + x . 2 1? x

( )

(d) Show that


[AQA June 1990]

1 2

0

tanh ?1 x dx = a ln b 2 , ( ) 1 ? x2

where a and b are numbers to be determined.

129

klmGCE Further Mathematics (6370)
y

Further Pure 2 (MFP2) Textbook

11. The diagram shows a region R in the x–y plane bounded by the curve y = sinh x, the x-axis and the line AB which is perpendicular to the x-axis. A

R

O (a) Given that AB = 3 , show that OB = ln 3. 4
2 (b) (i) Show that cosh ( ln k ) = k + 1 . 2k

B

x

(ii) Show that the area of the region R is 2 . 3

(c) (i) Show that



ln 3 0

sinh 2 x dx = 1 ? sinh ( ln 9 ) ? ln 9 ? ?. 4?

(ii) Hence find, correct to three significant figures, the volume swept out when the region R is rotated through an angle of 2 π radians about the x-axis.
[NEAB June 1998]

130

klmGCE Further Mathematics (6370)
7.1 7.2 7.3 Introduction Arc length Area of surface of revolution

Further Pure 2 (MFP2) Textbook

Chapter 7: Arc Length and Area of Surface of Revolution

This chapter introduces formulae which allow calculations concerning curves. When you have completed it, you will: ? ? ? know a formula which can be used to evaluate the length of an arc when the equation of the curve is given in Cartesian form; know a formula which can be used to evaluate the length of an arc when the equation of the curve is given in parametric form; know methods of evaluating a curved surface area of revolution when the equation of the curve is given in Cartesian or in parametric form.

131

klmGCE Further Mathematics (6370)
7.1 Introduction
You will probably already be familiar with some formulae to do with the arc length of a curve and the area of surface of revolution. For example, the area under the curve y = f ( x ) above the x-axis and between the lines x = a and x = b is given by A =

Further Pure 2 (MFP2) Textbook
y

∫a

b

y dx.

O

a

b

x

You will also be familiar with the formula V =

of revolution when that part of the curve between the lines x = a and x = b is rotated about the x-axis. The formulae to be introduced in this chapter should be committed to memory. You should also realise that, as with many problems, the skills needed to solve them do not concern the formulae themselves but involve integration, differentiation and manipulation of algebraic, trigonometric and hyperbolic functions – many of which have been introduced in earlier chapters.



b a

πy 2 dx which gives the volume of the solid

132

klmGCE Further Mathematics (6370)
7.2 Arc length

Further Pure 2 (MFP2) Textbook

The arc length of a curve is the actual distance you would cover if you travelled along it. In the diagram alongside, δ s is the arc length between the points P and Q on the curve y = f ( x ) . If P has the coordinates ( x, y ) , Q has coordinates

y P y O a δs δx

Q δy N y b x

( x + δ x, y + δ y )

and PN is parallel to the x-axis so that angle

PNQ is 90?, it follows that PN = δ x and QN = δ y.

Now, if P and Q are fairly close to each other then the arc length δ s must be quite short and PQN be approximately a right-angled triangle. Using Pythagorus’ theorem,

(δ s )
Dividing by (δ x ) ,
2

2

≈ (δ x ) + (δ y ) .
2 2 2

( )
δs δx
ds dx

2

?δ y ? ≈ 1+ ? ? . ?δx? ? dy ? = 1+ ? ? ? dx ?
2

In the limit as δ x → 0,

( )

2

ds = 1 + ? dy ? . ? dx ? dx ? ? Thus, s=

2



b a

? dy ? 1 + ? ? dx. ? dx ?

2

If y = f ( x ) , the length of the arc of curve from the point where x = a to the point where x = b is given by s=



b

a

? dy ? 1 + ? ? dx ? dx ?

2

133

klmGCE Further Mathematics (6370)
(δ s )
Dividing by (δ t ) ,
2 2

Further Pure 2 (MFP2) Textbook

A corresponding formula for curves given in terms of a parameter can also be derived. Suppose that x and y are both functions of a parameter t. As before,
≈ (δ x ) + (δ y ) .
2 2

As δ x → 0, δ t → 0 and

( ) ( ) ( ddst ) = ( ddxt ) + ??? ddyt ??? ds = dx + ? dy ? ( dt ) ?? dt ?? dt ? dy ? dx s= ∫ ( dt ) + ?? dt ?? dt,
δs δt
2

≈ δx δt

2

?δ y? +? ? . ? δt ?
2

2

2

2

2

2

t2

2

2

t1

where t1 and t2 are the values of the parameter at each end of the arc length being considered. The length of arc of a curve in terms of a parameter t is given by
s=

∫ ( )
t2 t1

dx dt

2

? dy ? + ? ? dt, ? dt ?

2

where t1 and t2 are the values of the parameter at each end of the arc The use of these formulae will be demonstrated through some worked examples.

134

klmGCE Further Mathematics (6370)
Example 7.2.1

Further Pure 2 (MFP2) Textbook

Find the length of the curve y = cosh x between the points where x = 0 and x = 2.

Solution
y = cosh x dy = sinh x. dx Therefore,
? dy ? 1 + ? ? = 1 + sinh 2 x ? dx ? = cosh 2 x ? dy ? 1 + ? ? = cosh x. ? dx ?
2 2

Now,

s=

∫ = cosh x dx ∫
0 2 0

2

? dy ? 1 + ? ? dx ? dx ?

2

= [sinh x ]0

2

= sinh 2 ? sinh 0 = sinh 2.

135

klmGCE Further Mathematics (6370)
Example 7.2.2
y = a (1 ? cos θ ) between θ = 0 and θ = 2 π is 8a.

Further Pure 2 (MFP2) Textbook

Show that the length of the curve (called a cycloid) given by the equations x = a (θ ? sin θ ) ,

Solution
dx = a 1 ? cos θ ( ) dθ dy = a sin θ . dθ Therefore,

( )
dx dθ

2

2 ? dy ? +? = a 2 (1 ? cos θ ) + a 2 sin 2 θ ? ? dθ ?

2

= a 2 1 ? 2 cos θ + cos 2 θ + sin 2 θ = a 2 (1 ? 2 cos θ + 1) = a 2 ( 2 ? 2 cos θ ) = 2a 2 (1 ? cos θ ) = 2a 2 2sin 2 θ 2

(

)
2 2

(using sin θ + cos θ = 1)

(using 2 sin θ = 1 ? cos 2θ )

2

( )
dx dθ Now,

2

? dy ? 2 2θ +? ? = 4a sin 2 θ d ? ? = 2a sin θ . 2 s=

2

∫ ( ) = 2a sin θ dθ ∫ 2 = ? 2a ?2cos θ ? ? ( 2)?
2π 0

dx dθ

2

? dy ? +? ? dθ ? dθ ?

2



0



= ?4a cos π ? ( ?4a cos 0 ) = 8a.

0

136

klmGCE Further Mathematics (6370)
7.3 Area of surface of revolution

Further Pure 2 (MFP2) Textbook

If an arc of a curve is rotated about an axis, it forms a surface. The area of this surface is known as the ‘curved surface area’ or ‘area of surface of revolution’. Suppose two closely spaced points, P and Q, are taken on the curve y = f ( x ) . This arc is rotated about the x-axis by 2 π radians. The coordinates of P and Q are ( x, y ) and
y P y O a b δs δx y x Q δy

( x + δ x, y + δ y ) , respectively, and the length of arc PQ is δs.
You can see from the diagram that the curved surface generated by the rotation is larger than that of the cylinder of width δs obtained by rotating the point P about the x-axis, but smaller than the area of the cylinder width δs obtained by rotating the point Q about the same axis. Using the formula S = 2 πrh for the area of the curved surface of a cylinder, the area of the former is 2 πyδ s and that of the latter is 2 π ( y + δ y ) δ s. If the actual area generated by the rotation of arc PQ about the x-axis is denoted by δA, it follows that 2 πyδ s < δ A < 2 π ( y + δ y ) δ s

2πy < δ A < 2π ( y + δ y ) . δs Now as δ x → 0, δ y → 0 and δ s → 0 so that the right-hand side of the inequality tends to 2πy. Therefore, dA = 2 πy ds
A= =

or, dividing by δs,

∫ ∫

b a

2 πy ds
2

? dy ? 2 πy 1 + ? ? dx a ? dx ?

b

(from section 6.2)

The area of surface of revolution obtained by rotating an arc of the curve y = f ( x ) through 2π radians about the x-axis between the points where x = a and x = b is given by A=



? dy ? 2 πy 1 + ? ? dx a ? dx ?

b

2

137

klmGCE Further Mathematics (6370)
Example 7.3.1

Further Pure 2 (MFP2) Textbook

Find the area of surface of revolution when the curve y = cosh x is rotated through 2π radians about the x-axis.

Solution
This was the curve used in example 6.2.1 – from there ? dy ? 1 + ? ? = cosh x. ? dx ? Hence,
A=
2

∫ = ∫

? dy ? 2 πy 1 + ? ? dx 0 ? dx ?
2 0 2 2

2

2

2 π cosh x cosh x dx

∫ cosh x dx = 2 π 1 (1 + cosh 2 x ) dx ∫2
= 2π
0 2 0

= π ? x + sinh 2 x ? ? 2 ? ? ?0

2

= π ? 2 + 1 sinh 4 ? ? π ?0 + 1 sinh 0 ? ? ? ? ? 2 2 ? ? ? ? = π ? 2 + 1 sinh 4 ? . ? ? 2 ? ?

Example 7.3.2
Show that the area of surface of revolution when the cycloid curve given by the equations x = a (θ ? sin θ ) , y = a (1 ? cos θ ) is rotated through 2π radians about the x-axis is 64 πa 2 . 3

138

klmGCE Further Mathematics (6370)
Solution
This was the curve used in example 6.2.2 – from there

Further Pure 2 (MFP2) Textbook

( )
dx dθ Now, A=

2

? dy ? +? = 2a sin θ . ? 2 ? dθ ?

2

∫ = ∫

2π 0 2π

2πy

( )
dx dθ

2

? dy ? +? ? dθ ? dθ ?

2

∫ {1 ? ( 2 cos 2 ?1)}sin 2 dθ θ θ θ = 4 πa ∫ ( 2sin 2 ? 2 cos 2 sin 2 ) dθ sin θ ? cos θ sin θ ) dθ . = 8πa ( ∫ 2 2 2
= 4πa 2
2 2π 2

0

2πa (1 ? cos θ ) 2a sin θ dθ 2

θ

θ

(using cos 2 x = 2 cos 2 x ? 1)

0 2π

2

2

0 2π

2

0

Now, consider

du = ? 1 sin θ dθ 2 2 and the integral becomes 3 u 2 ( ?2du ) = ? 2u 3

∫ cos

2

θ sin θ dθ . The substitution u = cos θ gives
2 2 2



=? Integrating for A,

2 cos3 θ 2 3

.


A = 8πa 2 ? ?2 cos θ + 2 cos3 θ ? ? ?0 2 3 2? ?

3 ?? ? ? = 8πa 2 ? ? ?2 cos π + 2 cos3 π ? ? ? ?2 cos 0 + 2 cos 0 ? ? ? ? ? 3 3 ?? ? ?? = 8πa 2 ? 2 ? 2 + 2 ? 2 ? ? 3 3? ? ? = 64 πa 2 . 3

139

klmGCE Further Mathematics (6370)
Miscellaneous exercises 7
1. (a) Show that (i) (ii) d tan θ = sech 2θ , ( ) dθ d = sechθ = ?sechθ tanh θ . ( ) dθ

Further Pure 2 (MFP2) Textbook

(b) A curve C is given parametrically by

x = θ ? tanh θ ,
(i) Show that

y = sechθ ,

θ ≥ 0.

( )
dx dθ

2

? dy ? 2 +? ? = tanh θ . θ d ? ?

2

(ii) The length of arc C measured from the point ( 0,1) to a general point with parameter θ is s. Find s in terms of θ and deduce that, for any point on curve, y = e ? s .
[AQA Specimen]

2. (a) Using only the definitions of cosh x and sinh x in terms of exponentials, (i) determine the exact values of cosh α and sinh α , where α = ln 9 , 4 (ii) establish the identities cosh 2 x ≡ 2 cosh 2 2 x ? 1 sinh 2 x ≡ 2sinh x cosh x. (b) The arc of the curve with equation y = cosh x, between the points where x = 0 and x = ln 9 , is rotated through one full turn about the x-axis to form a surface of 4 revolution with area S. Show that S = π ln 9 + p 4 for some rational number p whose value you should state.
[AEB June 2000]

(

)

140

klmGCE Further Mathematics (6370)
3. (a) (i) Using only the definitions prove the identity
cosh 2 θ ? sinh 2 θ = 1.

Further Pure 2 (MFP2) Textbook

cosh θ = 1 eθ + e?θ and sinh θ = 1 eθ ? e ?θ , 2 2

(

)

(

)

(ii) Deduce a relationship between sechθ and tanh θ . (b) A curve C has parametric representation x = sechθ , (i) Show that y = tanh θ .

( )
dx dθ

2

? dy ? 2 +? ? = sech θ . θ d ? ?

2

(ii) The arc of the curve between the points where θ = 0 and θ = ln 7 is rotated through one full turn about the x-axis. Show that the area of the surface generated is 36 π square units. 25
[AEB June 1997]

4. A curve has parametric representation

x = θ + sin θ ,
(a) Prove that dx dθ

y = 1 + cos θ ,

0 ≤ θ ≤ 2 π.

( )

2

? dy ? 2θ +? ? = 4 cos 2 . θ d ? ?

2

(b) The arc of this curve, between the points when θ = 0 and θ = π is rotated 2 about the x-axis through 2 π radians. The area of the surface generated is denoted by S. Determine the value of the constant k for which S =k
[AEB June 1996]

θ θ ∫ (1 ? sin 2 ) cos 2 dθ ,
π 2 2 0

and hence evaluate S exactly.

141

klmGCE Further Mathematics (6370)
5. The curve C is defined parametrically by the equations x = 1 t 3 ? t, y = t 2 , 3 where t is a parameter. (a) Show that dx dt

Further Pure 2 (MFP2) Textbook

( )

2

2 ? dy ? + ? ? = t2 +1 . ? dt ?

2

(

)

(b) The arc of C between the points where t = 0 and t = 3 is denoted by L. Determine (i) the length of L, (ii) the area of the surface generated when L is rotated through 2 π radians about the x-axis
[AEB January 1998]

6. (a) Given that a is a positive constant and that y = a 2 sinh ?1 x + x a 2 + x 2 , a use differentiation to determine the value of the constant k for which
dy = k a 2 + x2 . dx

()

(b) A curve has the equation y = sinh ?1 x + x 1 + x 2 . The length of the arc of curve between the points where x = 0 and x = 1 is denoted by L. Show that L = 5ln 5 + 12 . 8
[AEB January 2000]

142

klmGCE Further Mathematics (6370)
Chapter 1
Exercise 1A
1. (a) 2, ? π 4 (b) 3, π 2 (c) 4, π (c) 7.07, –1.71

Further Pure 2 (MFP2) Textbook

Answers to Exercises – Further Pure 2

(d) 2, ? 5π 6

2. (a) 3.16, 2.82

(b) 5, 0.93

Exercise 1B
1. (a) 2 cos ? π + i sin ? π 4 4

(

)

(c) 4 ( cos π + i sin π ) 2. (a) ? 2 + 2 i (b) ?2 ? 2 3 i

(b) 3 cos π + sin π 2 2 (d) 2 cos ? 5π + i sin ? 5π 6 6

( (

)

)

Exercise 1C
1. (a) 3 + i , 4 + 3i (b) ?1 + 8i , ? 14 + 2 i

Exercise 1D
1. (a) 1 ( 6 + 8i ) 5 (b) 2 + 2i

Exercise 1E
1. (a) 2 cos π + i sin π 3 2 2

(

)

(b)

z z1 ?z ? = 1 and arg ? 1 ? = arg z1 ? arg z2 z2 z2 ? z2 ?

143

klmGCE Further Mathematics (6370)
Exercise 1F
π? 1. (a) ? ?6, ? ? 2? 3 5π ? (b) ? ? , ? ?2 6 ? 2 ?5 π ? (e) ? ? , ? ?9 6 ? 4π ? (c) ? ?9, ? 3 ? ?

Further Pure 2 (MFP2) Textbook

(d)

( 27 , 0 )

2. (a) 2.5 + 0.5i

(b) 4 + 2i

(c) 1 ? 1 i 3

Exercise 1G
1. (a) 5.39, 0.38 (b) 5, 0.93 (c) 7.62, –1.98

Exercise 1H
1. (a) y (b) y
π 4

(c)

y

3

( 2,1)

O

x

O

(1,0 )

x

O

x

y

2. (a)

y

(b)

( 0, 3)
( 4,2 )

O

x y

O

x

y 3.
O

P

( 0,1)

(1,1)

4. x

( ?1, ?2 )

x
( ?3,0)

Q

The locus is the line PQ

144

klmGCE Further Mathematics (6370)
Miscellaneous exercises 1
1. 3 ? 2i 2. (a) 1 ( 4 + 3i ) 5 3. 1 ? 2i, 4. 1 (1 ? i ) 5 5. (a) y Q
3,1

Further Pure 2 (MFP2) Textbook

y (b)
O

z

4 ? 2i

z*

( ) x ( )
4,3 5 5 4 ,? 3 5 5

(c) 1, 1.2

P

(b) 3

(c) π 3

O 6. y
2, 3

3, 0

x

arg z = 1 π 4
z ? 2 ? 3i = 3

O

2, 0

x 2, 3π 4 (b)(ii)
y O P1

7. (a)(i) ?1 + i (ii)

( )
4, π 6

(iii) 2 10

(
8. (a) y

2 2, ? 7 π 12

)

x

P2

(b)

y
A

?1, 0

x

?3, 0 ?1, 0

x

9. (a) 14 + 2i, 1+i

(b)(i)

20, 0.46 and

10, ? 0.32 (ii)

10

145

klmGCE Further Mathematics (6370)
Chapter 2
Exercise 2A
1. (a) ?3 ± i (b) ?5 ± i

Further Pure 2 (MFP2) Textbook

Exercise 2B
1. (a) ?1, ? 1, 3 (b) 1, 1 ± i (c) 2, ? 2 ± i

Exercise 2C
1. (a) 7, 12, –5 (b) ?4 , ?7 , ?2 3 3 3

2. 2 x3 ? 6 x 2 + 7 x + 10 = 0

Exercise 2D
1&2 (a) x3 ? 3 x 2 ? 36 x ? 189 = 0 3. (a) 2 x3 ? 9 x 2 + 162 = 0 (b) x3 ? 4 x 2 + x ? 5 = 0 (b) 2 x3 ? 9 x 2 + 12 x + 1 = 0 (c) 7 x3 + 8 x 2 + 4 x ? 8 = 0

Exercise 2E
1. (a) 16 x 4 ? 6 x 2 + 5 x ? 4 = 0 (b) 3

Exercise 2F
1. x3 ? 4 x 2 + 6 x ? 4 = 0 2. 2, –3i 3. 1 + i, ± i 2

146

klmGCE Further Mathematics (6370)
Miscellaneous exercises 2
1. (a) α = 1, β = 5 (b) ?2 2. (b) (i) β = 3 ? 4i, γ = ?6 (ii) ?150 (iii) 0, – 11, 150 3. (a)

Further Pure 2 (MFP2) Textbook

∑ α = 0 ∑ αβ = 3 2

αβγ = ?2

(b) 2 x3 ? 3 x 2 ? 8 = 0 4. (a) 8 7 (b) 1 ? 2i , ?6 7 5. (a) p = ?3 (b) (i) q = 7 (ii)

∑α 2 < 0

(c) (i) – 3 (ii) 75 6. (b) (i) p = 4, q = ?4

147

klmGCE Further Mathematics (6370)
Chapter 33 Chapter
Exercise 3A
1. (a) 2r (b) 1 n ( n + 1) 2

Further Pure 2 (MFP2) Textbook

1 2. (b) 1 ? 18 3 ( n + 1)( n + 2 )( n + 3)

3. (b) 1 n ( n + 1)( n + 2 ) 6

Miscellaneous exercises 3
2. (b) n ( n + 1) n 2 + n + 1 9. (a) 2 10. (a) 1 (c) 1

(

)

148

klmGCE Further Mathematics (6370)
Chapter 4
Exercise 4A
2. (a) cos15θ + i sin15θ (e) ?64 (b) 1 (f) ? 1 64

Further Pure 2 (MFP2) Textbook

(

3 +i

)

(c) i (g) ?41472 3

(d) ?8i

Exercise 4B
1. 3sin θ ? 4sin 3 θ
3 θ 2. 3 tan θ ? tan 1 ? 3 tan 2 θ

3. 16sin 5 θ ? 20sin 3 θ + 5sin θ

Exercise 4C
? 1. (a) 1 ? z4 + 1 4 ? 2? z ? ? ? (b) 1 ? z7 + 1 7 ? 2? z ? ? ? (c) 1 ? z6 ? 1 6 ? 2i ? z ? ? ? (d) 1 ? z3 ? 1 3? 2i ? z ? ?

Exercise 4D
1. (a) 2e 4
πi

(b) 2e

? πi 6

(c) 12e 6

π

(d) 4e 6



Exercise 4E
1. (a) –1 (b) 7 (c) 8

Exercise 4F
1. (a) ±1, ± i (b) 2 cos 2kπ + i sin 2kπ , 5 5 k = 0 ± 1, ± 2, ± 3, ± 4, 5 k = ±1, ± 2

(

)

k = 0 ± 1, ± 2

(c) cos kπ + i sin kπ , 5 5 2. cos 2kπ + i sin 2kπ , 5 5 3. 1 , ± 1 i 2 2

149

klmGCE Further Mathematics (6370)
Exercise 4G
(1+ 2 k ) πi
1 ( 8 k ?1) πi 2 6 e 12

Further Pure 2 (MFP2) Textbook

1. (a) 2e (c)

8

k = 0, 1, 2, 3

(b)

k = 1, 2, 3

1 ( 6 k ?1) πi 2 8 e 24

k = 1, 2, 3, % , 8

(d) ± 1 ( i ? 1) 2 (f) ? 1 cot kπ ? i 2 5

( 4 k +1) πi

(e) 2e

6

? 1 k = 0, 1, 2

(

) )
k = 0, ± 1, 2

Miscellaneous exercises 4
1. (a) 64, +π (b) 2 2 cos π + i sin π , 2 cos ? π + i sin ? π 4 4 6 6 (c) ±2 (1 + i ) , ± 2 (1 ? i )

(

) (

2. (a)

2 cos π + i sin π , 2 cos? π + i sin ? π 4 4 6 6 (b)(ii) 1.41 + 0.12i, ? 0.81 + 1.16i, ? 0.60 ? 1.28i (c) 5? 5 , 8 5+ 5 8
y
πi(1+2k ) 5

(

) (

)

3. (b) ± 5 ± 5 8 4. (b) 1, 3 5. (b) z = 1 + e (d)(i) π 10

k = ±1, ± 2 (ii) 2 cos π 10

(c) centre z = 1, radius 1

1

x

6. (b)(ii) x 2 + x ? 1 = 0 7. (a)(ii) 2i sin nθ 8. (a)(i) 2sin θ 2

(iii) 2 cos 2π , 2 cos 4π 5 5 (b)(i) A = 1, B = ?3 (b) e
kπi 3

(iv) ?1 + 5 , 4

?1 ? 5 4

k = ±1, ± 2

(c)(i) coefficients of w6 cancel
3πi 2e 4 , ? 7πi 2e 12

(ii) ? 1 2
y
2

9. (b)

(c)

B

A
2

(d) 3 3 2
x

2

C

2

150

klmGCE Further Mathematics (6370)
Chapter 5
Exercise 5A
1. (a) π 4 (b) π 6 (c) ? π 6 (d) π 2

Further Pure 2 (MFP2) Textbook

(e) ? π 6

(f) π

Exercise 5C
1. (a) 3 1 + 9 x2 x + tan ?1 x 1 + x2 2e x 1 ? 4 x2 2 x2 ?8 + 12 x ? 4 x 2 (b) ?3 6x ? 9x
2

(c)

2 1 ? 4 x2

2. (a)

(b) e x cos ?1 2 x ?

(c) 2 x sin ?1 ( 2 x ? 3) + 3

3. (a)

x3

?1 ? 3sin 4 3 x x 1 ? 9 x2

(b)

?1 + 3x (1 + x ) ? ( ?
2

6x

2

2 ? +1 ? ?

)

?

2 x tan ?1 3 x 2 + 1

(1 + x )
2

(

)

2

4. (a)

a 1 ? ( ax + b )
2

(b)

a 2 1 + ( ax + b )

151

klmGCE Further Mathematics (6370)
Exercise 5D
1. π 6 2. π

Further Pure 2 (MFP2) Textbook

3. π [note that sin ?1 4 + sin ?1 3 = π by drawing a 3, 4, 5 right-angled triangle]. 2 5 5 2 4. π 4 5. π 2

Exercise 5E
1. (a) tan ?1 ( x + 2 ) (b) 1 tan ?1 2 x ? 1 ( ) 3 6 (c) 2 tan ?1 2 x ? 1 7 7

? ? 2. (a) ln x 2 + 2 x + 3 ? 2 tan ?1 ? x + 1 ? ? 2 ?

(

)

? ? (b) 1 ln x 2 + x + 1 ? 1 tan ?1 ? 2 x + 1 ? 2 3 ? 3 ?

(

)

3. (a) sin ?1 x ? 3 4

( )

(b) sin ?1 x ? 1 2

( )

(c)

1 sin ?1 4 x ? 1 ( ) 2

4. (a) sin ?1 x ? 1 ? x 2 (b) ?3 3 + 2 x ? x 2 + sin ?1 x ? 1 2 (c)

( )

? ? 1 ? x ? x 2 + 3 sin ?1 ? 2 x + 1 ? 2 ? 5 ?

152

klmGCE Further Mathematics (6370)
Chapter 6
Exercise 6A
1. (a) 2 x e + e? x
2x (b) e 2 x + 1 e ?1
x (c) e x ? 1 e +1

Further Pure 2 (MFP2) Textbook

(d)

2 e ? e?3 x
3x

Exercise 6B
1. (a) 0.64 (b) 0.86 (c) 0.24 (d) –0.54 (e) 1.05 (f) 1.00

Exercise 6C
1. (a)
y

y

1
O y x

(b)

O

x

(c)

1

O

x

–1

Exercise 6E
2. 8cosh 4 x ? 8cosh 2 x + 1

Exercise 6F
1. (a) 3sinh 3 x (d) 2 x sinh 2 x2? cosh 2 x x (g) ?cosech x coth x (b) 6sinh 3 x cosh 3 x (e) tanh x + x sech 2 x (c) 2 x cosh x + x 2 sinh x (f) ?sech x tanh x

153

klmGCE Further Mathematics (6370)
Exercise 6G
1. (a) 1 sinh 3 x + c 3 (c) 1 x cosh 2 x ? 1 sinh 2 x + c 2 4

Further Pure 2 (MFP2) Textbook

(b) 1 x + 1 sinh 2 x + c 2 2 (d) x ? tanh x

(

)

Exercise 6H
3. (a) ln 1 3 + 5 2

(

)
(b) e
x

(b) 1 ln 2 2

(c) ln1.5

Exercise 6I
1. (a) 3 9 ? x2 1 9+ x
2

(c) (e)

1 x ? 16 2 ? 1 cosh ?1 x 2 2 4 x ?1 x
2

(d) e x sinh ?1 x +

1 + x2

2. 2 3 + ln 2 + 3 3

(

)
(b) cosh ?1 x + c 4 ?1 x + 1 (e) sinh +c 2 (h) cosh ?1 x ? 1 + c 3 (c) 1 sinh ?1 2 x + c 2 5 ?1 x ? 2 (f) cosh +c 4

Exercise 6J
1. (a) sinh ?1 x + c 3 1 ?1 3 x (d) sinh +c 3 7 (g) sinh ?1 ( x + 2 ) + c

Exercise 6K
1. (a) ln 2 (d) ln 5 + 21 2 (b) 1, ? ln 7 (e) 0 (c) ln

(

2 ? 1 , ln 4 + 15

) (

)

154

klmGCE Further Mathematics (6370)
Miscellaneous exercises 6
1. (a) e x 2. 1 ln 2 2 3. (b) ? ln 2, ln 2 + 5 4. (b) ln 4 5. (c) ? 2 , ? 4 3 3 7. (b)(i) 1 tanh 2 x 2 8.
y

Further Pure 2 (MFP2) Textbook

(b) 1

(

)

(ii) ln cosh x ? 1 tanh 2 x 2 (b)(i) ln 3
two roots one root no roots
x

(ii) ln 3 4

1
O

9. (a) x ≥ 0
2 10. (d) 1 ( ln 3) 8

(b)(ii)

2 ? ln

(

2 +1

)

11. (c)(ii) 1.76

155

klmGCE Further Mathematics (6370)
Chapter 7
Miscellaneous exercises 7
1. 2. 3. 4. 5. 6. (b)(ii) s = ln cosh θ (a)(i) 97 , 72 65 72 (b) 6305 5184

Further Pure 2 (MFP2) Textbook

(a)(ii) 1 ? tanh 2 θ = sech 2θ (b) k = 8π, (b)(i) 12 (a) 2 20 2 π 3 (ii) 576 π 5

156


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