P1: FXS/ABE P2: FXS CHAPTER 16 - Cambridge University Press · P1: FXS/ABE P2: FXS 9780521740494c16a.xml CUAU033-EVANS September 12, 2008 10:50 414 Essential Advanced General Mathematics

P1: FXS/ABE P2: FXS

9780521740494c16a.xml CUAU033-EVANS September 12, 2008 10:50

C H A P T E R

16Polar coordinates and

complex numbers

ObjectivesTo describe points on the plane using polar coordinates

To describe graphs with polar coordinates

To transform polar coordinates to cartesian coordinates

To transform cartesian coordinates to polar coordinates

To understand the imaginary number i

To understand the set of complex numbers C

To understand the real-valued functions of the complex numbers, Re(z) and Im(z)

To represent complex numbers graphically on an Argand diagram

To understand the rules which define equality, addition, subtraction and

multiplication of complex numbers

To understand the concept of the complex conjugate

To understand the operation of division by complex numbers

To understand the modulus-argument form of a complex number and the basic

operations on complex numbers in that form

To understand the geometrical significance of multiplication and division of

complex numbers in the modulus-argument form

To be able to factorise quadratic polynomials over C

To be able to solve quadratic polynomials over C

A new set of numbers called Complex numbers is introduced in this chapter. The need for this

new set of numbers can be equated to the need for a solution of the equation x2 + 1 = 0. A

geometric interpretation is also shown to be useful.

Complex numbers can be expressed in two ways, cartesian form and polar form. As a

preliminary to this, polar coordinates are introduced.

410

SAMPLE

Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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Chapter 16 — Polar coordinates and complex numbers 411

16.1 Polar coordinatesIn previous work the cartesian coordinate system has been used to represent points in

two-dimensional space. The point (x, y) is described in terms of its horizontal displacement (x)

and its vertical displacement (y) from a fixed point called the origin (O).

An alternative way of locating the point P is to describe it in terms of its polar coordinates

[r, �] where r specifies the distance from the origin or pole and � specifies the angle of the line

OP relative to the line OZ which extends to the right from O and is called the polar axis.

Note: An angle in an anticlockwise direction from OZ is

considered to be positive.

θO Z

r

P[r, θ]

For example, the point P[4, 60◦] is located a

distance of 4 units along a line forming an angle

of 60◦ with the polar axis.60°

O Z

4

P[4, 60°]

Using this system it is clear that any point can be specified in a number of different ways.

For example, the point [4, 60◦] may also be specified by [−4, −120◦].

The angle � = 120◦ is measured in a clockwise direction from O.

The diagram below and to the left illustrates the point P ′[4, −120◦] and the diagram to the

right, [−4, −120◦].

P'

4120°

OZ

60°

4

120°

O Z

P

P[4, 60◦] may also be specified by P[4, −300◦] or P[−4, 240◦].

Example 1

Plot the point P with coordinates [3, −30◦].

Solution

3

30°O Z

P

SAMPLE


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412 Essential Advanced General Mathematics

The relationship between cartesian andpolar coordinatesIf a set of cartesian axes is superimposed

over a polar axis, the relationship between

cartesian and polar coordinates can be

established.

x(Z)

y

Oθ

r

r cos(θ)

r sin(θ)

P(x, y)

From the diagram

x = r cos � 1

and y = r sin � 2

The angle � can be found by finding a solution which satisfies both equations 1 and 2

Squaring both sides of equations 1 and 2 and adding yields

x2 + y2 = r2 cos2 � + r2 sin2 �

= r2(cos2 � + sin2 �)

i.e. x2 + y2 = r2

Using these relationships, coordinates can be converted from cartesian to polar and vice versa.

Example 2

a Express (√

3, 1) in polar form. b Express [√

2, 45◦] in cartesian form.

Solution

a r2 = x2 + y2 sin � = 1

2= (

√3)2 + (1)2

= 4 and cos � =√

3

2∴ r = 2 ∴ � = 30◦

∴ (√

3, 1) specifies the same point as [2, 30◦]

b r =√

2 and � = 45◦

x =√

2 cos 45◦ =√

2 × 1√2

= 1

y =√

2 sin 45◦ =√

2 × 1√2

= 1

∴ [√

2, 45◦] specifies the same point as (1, 1)SAMPLE


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Curve sketching using polar coordinatesIn the same way that graphs of relationships in cartesian form can be sketched, relationships

expressed in polar form can also be sketched. Some very interesting curves result from simple

polar equations.

It is recommended that to sketch these graphs a graphics calculator or a computer graphing

package be used. Graphs can of course be plotted using a table of values. A sheet of polar

graph paper is also useful although a sheet of blank paper will suffice as long as a ruler and a

protractor are used.

Using the TI-NspireExample 3

Plot the graph of r = 3 (1 − cos �).

Solution

Open a Graphs & Geometry application

(c 2) and select Polar from the GraphType menu (b 33).

Enter r1 (�) = 3 (1 − cos (�)) as shown

Note that the domain of ( as well as the

step size can be adjusted in this window.

The graph is shown using the Zoom,Fit command from the Window menu

(b 4 ).

Note that every point on the graph satisfies

r = 3 (1 − cos (�)) . For example, for

� = 60◦

r = 3 (1 − cos (60◦)) = 3

(1 − 1

2

)= 3

2For � = 180◦, r = 3 (1 − cos (180◦))

= 3 (1 − (−1)) = 6

For � = −90◦, r = 3 (1 − cos (−90)) = 3◦

SAMPLE


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Note that Trace (b 5) can be used to

show the coordinates of the points on the

graph in the form [r, �].

To go to the point where � = �, simply

type � followed by enter. The cursor will

then move to the point [r, �] = [6, �] as

shown.

Example 4

Sketch the graph of r = �.

Solution

Open a Graphs & Geometry application

(c 2) and select Polar from the GraphType menu (b 33).

Enter r1 (�) = �.

The graph is shown.

If the domain of � is extended, the graph

continues to spiral out. This can be

observed by extending the domain to

0 < � < 6�.

The resulting graph is shown using the

Zoom, Out command from the Windowmenu (b 44).SAM

PLE


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Using the Casio ClassPadPlot the graph of r = 3 (1 − cos �).

Ensure that the mode is set to radians.

In tap and from the menu select

.

Enter the equation r1 = 3 (1 − cos (�)) and tap

$ to produce the graph.

In the screen shown, the window was selected by tapping Zoom, quick initialize.

Example 5

Find the polar equation of the circle whose cartesian equation is

x2 + y2 = 4x

Solution

Let x = r cos � and y = r sin �

Then r2 cos2 � + r2 sin2 � = 4r cos �

∴ r2(cos2 � + sin2 �) = 4r cos �

∴ r2 − 4r cos � = 0

∴ r (r − 4 cos �) = 0

∴ r = 0 or r = 4 cos �

∴ r = 4 cos � is the polar equation of the circle

Example 6

Find the cartesian equation corresponding to each of the following polar equations.

a r = 3 b r = 1

1 + sin �c r = 3(1 − cos �)SAMPLE


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Solution

a r = 3√x2 + y2 = 3

x2 + y2 = 9The circle with centre (0, 0)

and radius 3

b r = 1

1 + sin �implies r (1 + sin �) = 1

i.e. r + r sin � = 1

∴√

x2 + y2 = 1 − y

x2 + y2 = 1 − 2y + y2

∴ x2 = −2

(y − 1

2

)

∴ y = − x2

2+ 1

2c r = 3(1 − cos �)

∴ r2 = 3r − 3r cos � (Multiplying both sides of equation by r )

∴ x2 + y2 = 3√

x2 + y2 − 3x

x2 + y2 + 3x = 3√

x2 + y2

Exercise 16A

1 Plot each of the following points using a polar axis.Example 1

a A[2, 30◦] b B[3, 45◦] c C[−2, 60◦] d D[4, −30◦]

e E[5, 50◦] f F[−5, −50◦] g G[−5, 130◦] h H [5, −130◦]

2 Plot each of the following points using a polar axis.

a A[1, �] b B[−1, −�] c C[2,

�

2

]d D

[3,

3�

4

]

3 Convert the following cartesian coordinates to polar coordinates. (Remember to note

which quadrant each point is in.)

Example 2a

a (4, 4) b (1, −√3) c (2

√3, −2) d (−5, 12)

e (6, −5) f (√

3, 1) g (−5, −12) h (4, 3)

4 Convert the following polar coordinates to cartesian coordinates.Example 2b

a [−2, 30◦] b[−4,

�

2

]c

[−1,

5�

4

]d [4, −2�]

e

[2, −7�

6

]f [5, 240◦] g [2, 180◦] h [1, −120◦]

5 Plot each of the following polar graphs.Examples 3, 4

a r = 3

cos �b r = 4

sin �c r = 2 cos �

d r = 2�, 0 ≤ � ≤ 6� e r = �

�,

�

6≤ � ≤ 4� f r = cos 2�

g r = 5(1 + cos �) h r = 2(1 − sin �) i r = 3 cos � + 2

j r = ±√cos 2� k r =

√�

�, 0 ≤ � ≤ 6� l r = 2 sin 2�

SAMPLE


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6 Obtain the polar equations of each of the following.Example 5

a x2 + y2 = 16 b x + y = 6 c x2 = y dx2

4+ y2 = 1

7 Obtain the cartesian equations of each of the following.Example 6

a r = 2 b r = a(1 + cos �) c r = a cos �

d r = 2a(1 + sin 2�) e r = a

1 + cos �f r = a

1 + sin �

16.2 The set of complex numbersIn earlier work in mathematics it was assumed that an equation of the form x2 = −1 had no

solutions. Mathematicians of the eighteenth century introduced the imaginary number i with

the property i2 = −1. i is defined as i = √−1 and the equation x2 = −1 has two solutions, i

and −i . By considering i such that i2 = −1 then the square roots of all negative numbers may

be found.

For example√−4 = √

4 × −1

=√

4 × √−1

= 2i

Imaginary numbers led to the introduction of complex numbers, which further broadened the

scope of mathematical thinking. Today complex numbers are widely used in engineering, the

study of aerodynamics and many other branches of physics.

Consider the equation x2 + 2x + 3 = 0. Using the quadratic formula to solve yields:

x = −2 ± √4 − 12

2

= −2 ± √−8

2= −1 ± √−2

This equation has no real solutions since the discriminant � = b2 − 4ac is less than zero.

However, for complex numbers

x = −1 ±√

2i

A complex number is an expression of the form a + bi , where a and b are real numbers. C

is the set of complex numbers, i.e. C = {a + bi : a, b ∈ R}. The letter often used to denote a

complex number is z.

Therefore z ∈ C implies z = a + bi where a, b ∈ R

If a = 0, z is said to be imaginary.

If b = 0, z is real.

Real numbers and imaginary numbers are subsets of C.SAMPLE


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Functions of complex numbersLet z = a + bi

Re (z) is a function which defines the real component of z. Im (z) is a function which defines

the value of the imaginary component of z.

Re (z) = a and Im (z) = b

Note: Re (z) and Im (z) are both real-valued functions of z, i.e. Re : C → R and Im : C → R.

So for the complex number z = 2 + 5i, Re (z) = 2 and Im (z) = 5.

Equality of complex numbersTwo complex numbers are equal if and only if both their real and imaginary parts are equal.

i.e. x1 + y1i = x2 + y2i

if and only if x1 = x2 and y1 = y2

Example 7

If 4 − 3i = 2a + bi find the values of a and b.

Solution

2a = 4 and b = −3

a = 2

Example 8

Find the values of a and b such that (2a + 3b) + (a − 2b)i = −1 + 3i

Solution2a + 3b = −1 1

a − 2b = 3 2

2 × 2 gives

2a − 4b = 6 3

1 − 3 gives

7b = −7

∴ b = −1 and a = 1

Operations with complex numbersAddition and subtractionIf z1 = a + bi and z2 = c + di (a, b, c, d ∈ R)

Then z1 ± z2 = (a ± c) + i(b ± d)

i.e. Re (z1 ± z2) = Re (z1) ± Re (z2) and Im (z1 ± z2) = Im (z1) ± Im (z2)

SAMPLE


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Example 9

If z1 = 2 − 3i and z2 = −4 + 5i find

a z1 + z2 b z1 − z2

Solution

a z1 + z2 = 2(2 + −4) + (−3 + 5)i

= −2 + 2i

b z1 − z2 = (2 − −4) + (−3 − 5)i

= 6 − 8i

Multiplication by a real constant

If z = a + bi and k ∈ R

then kz = k(a + bi)

= ka + kbi

For example, if z = 3 − 6i then 3z = 9 − 18i

Multiplication by powers of iSuccessive multiplication by powers of i gives the following:

i1 = i

i2 = −1

i3 = −i

i4 = (−1)2 = 1

i5 = i

and so onIn general, for n = 0, 1, 2, 3, . . .

i4n = 1

i4n+1 = i

i4n+2 = −1

i4n+3 = −i

When multiplying by powers of i, the usual index laws apply.

Example 10

Simplify

a i13 b 3i4 × (−2i)3

Solution

a i13 = i4×3+1

= i

b 3i4 × (−2i)3 = 3 × (−2)3 × i4 × i3

= −24i7

= 24i

SAMPLE


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Exercise 16B

1 State the values of Re (z) and Im (z) for each of the following.

a 2 + 3i b 4 + 5i c1

2− 3

2i

d −4 e 3i f√

2 − 2√

2i

2 Find the values of a and b in each of the following ifExamples 7, 8

a 2a − 3bi = 4 + 6i b a + b − 2abi = 5 − 12i

c 2a + bi = 10 d 3a + (a − b)i = 2 + i

3 Simplify the following.Example 9

a (2 − 3i) + (4 − 5i) b (4 + i) + (2 − 2i)

c (−3 − i) − (3 + i) d (2 − √2i) + (5 − √

8i)

e (1 − i) − (2i + 3) f (2 + i) − (−2 − i)

g 4(2 − 3i) − (2 − 8i) h −(5 − 4i) + (1 + 2i)

i 5(i + 4) + 3(2i − 7) j1

2(4 − 3i) − 3

2(2 − i)

4 SimplifyExample 10

a√−16 b 2

√−9 c√−2 d i3 e i14

f i20 g −2i × i3 h 4i4 × 3i2 i√

8i5 × √−2

5 Simplify

a i(2 − i) b i2(3 − 4i) c√

2i(i − √2) d −√

3(√−3 + √

2)

16.3 Multiplication and division ofcomplex numbersMultiplication of complex numbersIf z1 = a + bi and z2 = c + di(a, b, c, d ∈ R)

Then z1 × z2 = (a + bi) (c + di)

= ac + bci + adi + dbi2

= (ac − bd) + (bc + ad)i (bdi2 = −bd)

Example 11

If z1 = 3 − 2i and z2 = 1 + i , find z1z2.

Solution

z1z2 = (3 − 2i)(1 + i)

= 3 − 2i + 3i − 2i2

= 5 + i

SAMPLE


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Conjugate of a complex numberIf z = a + bi then the conjugate of z denoted by the symbol z is

z = a − bi

For example, the conjugate of −4 + 3i is −4 − 3i and vice versa.

Note that zz = (a + bi) (a − bi)

= a2 + abi − abi − b2i2

= a2 + b2 which is a real number

Using this result, a2 + b2 can now be factorised over the set of complex numbers.

Example 12

If z1 = 2 − 3i and z2 = −1 + 2i find

a (z1 + z2) and z1 + z2 b z1z2 and z1 z2

Solution

z1 = 2 + 3i and z2 = −1 − 2ia z1 + z2 = (2 − 3i) + (−1 + 2i)

= 1 − i

(z1 + z2) = 1 + i

z1 + z2 = 2 + 3i + −1 − 2i

= 1 + i

b z1z2 = (2 − 3i)(−1 + 2i) = 4 + 7i

z1z2 = 4 − 7i

z1 z2 = (2 + 3i)(−1 − 2i) = 4 − 7i

In general it can be stated that

the conjugate of the sum of two complex numbers is equal to the sum of the conjugates

the conjugate of the product of two complex numbers is equal to the product of the

conjugates.

i.e. (z1 + z2) = z1 + z2 and (z1z2) = z1 z2

Division of complex numbersDivision of one complex number by another relies on the fact that the product of a complex

number and its conjugate is a real number.

If z1 = a + bi and z2 = c + di (a, b, c, d ∈ R)

Thenz1

z2= (a + bi)

(c + di)

If the numerator and denominator are multiplied by the conjugate of z2 then

z1

z2= (a + bi)

(c + di)× (c − di)

(c − di)

= ac + bci − adi − bdi2

c2 + d2

= (ac + bd)

c2 + d2+ (bc − ad)i

c2 + d2

SAMPLE


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Example 13

If z1 = 2 − i and z2 = 3 + 2i , findz1

z2.

Solution

z1

z2= 2 − i

3 + 2i× 3 − 2i

3 − 2i

= 6 − 3i − 4i + 2i2

32 + 22

= 4 − 7i

13

= 1

13(4 − 7i)

Example 14

Solve for z the equation (2 + 3i) z = −1 − 2i

Solution

(2 + 3i)z = −1 − 2i

z = −1 − 2i

2 + 3i

= −1 − 2i

2 + 3i× 2 − 3i

2 − 3i

z = −8 − i

13

There is an obvious similarity in the process of expressing a complex number with a real

denominator and the process of rationalising the denominator of a surd expression.

Example 15

If z = 2 − 5i find z−1 and express with a real denominator.

Solution

z−1 = 1

z

= 1

2 − 5i

= 1

2 − 5i× 2 + 5i

2 + 5i

= 2 + 5i

29

= 1

29(2 + 5i)

SAMPLE


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Using the TI-NspireThe TI-Nspire can be used to deal with complex numbers. Select Rectangular form in

the Document Settings (c 81) to perform calculations on complex numbers in

the form a + bi .

The square root of a negative number can now

be performed as shown.

The results of the operations +, −, × and √,

are illustrated using the two complex numbers

2 + 3i and 3 + 4i .

It is possible to perform arithmetic

operations with complex numbers as shown.

The Real Part command from the ComplexNumber Tools submenu of the Number menu

(b 282) can be used as shown to find

the real part of a complex number.

The Magnitude command from the

Complex Number Tools submenu of the

Number menu (b 285) can be used as

shown to find the modulus of a complex number.

The Complex Conjugate command from the

Complex Number Tools submenu of the Numbermenu (b 281) can be used as shown

to find the complex conjugate of a complex

number.

SAMPLE


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There are also commands for factorising

polynomials over the complex numbers and for

solving polynomial equations over the complex

numbers. These are available from the Complexsubmenu of the Algebra menu (b 3 ).

Using the Casio ClassPadIn tap Real in the status bar at the bottom of the screen to enter Cplx mode. In this

mode enter√−1 and tap to obtain the answer i. Similarly,

√−16 will return the

answer 4i.

Operationsi is found in in the on-screen keyboard.

With the calculator set to Complex mode, a

number of arithmetic operations can be carried out,

as shown in the screen at right using options from

Interactive, Complex.Polynomials can be factorised and solved over the

complex number field using Interactive, transformationand Equation/inequality, solve.

Exercise 16C

1 Expand and simplifyExample 11

a (4 + i)2 b (2 − 2i)2 c (3 + 2i)(2 + 4i)

d (−1 − i)2 e (√

2 − √3i)(

√2 + √

3i) f (5 − 2i)(−2 + 3i)

SAMPLE


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2 Write down the conjugate of each of the following complex numbers.

a 2 − 5i b −1 + 3i c√

5 − 2i d −5i

3 If z1 = 2 − i and z2 = −3 + 2i findExample 12

a z1 b z2 c z1z2 d z1z2

e z1z2 f z1 + z2 g z1 + z2 h z1 + z2

4 If z = 2 − 4i express each of the following in the form x + yi .Example 15

a z b zz c z + z d z(z + z)

e z − z f i(z − z) g z−1 hz

i5 Find the values of a and b such that (a + bi)(2 + 5i) = 3 − i

6 Express in the form x + yiExample 13

a2 − i

4 + ib

3 + 2i

2 − 3ic

4 + 3i

1 + i

d2 − 2i

4ie

1

2 − 3if

i

2 + 6i

7 Find the values of a and b if (3 − i)(a + bi) = 6 − 7i

8 Solve each of the following for z.Example 14

a (2 − i)z = 4 + 2i b (1 + 3i)z = −2 − i c (3i + 5)z = 1 + i

d 2(4 − 7i)z = 5 + 2i e z(1 + i) = 4

16.4 Argand diagramsAn Argand diagram is a geometrical representation of the set of complex numbers. In a

vector sense, a complex number has two dimensions; the real part and the imaginary part.

Therefore a plane is required to represent C.

An Argand diagram is drawn with two

perpendicular axes. The horizontal axis

represents Re(z), z ∈ C , and the vertical

axis represents Im(z), z ∈ C .1

1

2

2

3

3–3

–3

–2

–2

–1

–1

0

(–2 + i) (3 + i)

(2 – 3i)

Re(z)

Im(z)

Each point on an Argand diagram represents

a complex number. The complex number a + bi

is situated at the point (a, b) on the equivalent

cartesian axes as shown by the examples in this

figure. The number written as a + bi is

called the cartesian form of the complex number.SAMPLE


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Example 16

Write down the complex number represented

by each of the points A to F on this Argand diagram.45

5

–5

–5 0 Re(z)

Im(z)

AB

CD

FE

Solution

A : 2 + 3i B : 4i

C : −5 D : −1 + i

E : −5 − 2i F : 1 − 3i

Geometrical representation of the basicoperations on complex numbersThe addition of two complex numbers is similar to a vector sum and follows the triangle of

vectors rule.

The multiplication by a scalar follows vector properties of parallel position vectors.

Re(z)

Im(z)

0

z1 + z2

z1

z2

Re(z)

Im(z)

0bz

cz

az

z

a > 1

0 < b < 1

c < 0

The subtraction z1 − z2 is represented by the sum z1 + (−z2).

Example 17

a Represent the following points on an Argand diagram.

i 2 ii −3i iii 2 − i iv −(2 + 3i) v −1 + 2i

b Let z1 = 2 + i and z2 = −1 + 3i .

Represent z1, z2, z1 + z2 and z1 − z2 on an Argand diagram and verify that the complex

number sum and difference follow the vector triangle properties.

SAMPLE


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Solution

a

1

1

2

2

2

3

3–3

–3

–2

–2

–1–1

0

–(2 + 3i)

–1 + 2i

2 – i

–3i

Re(z)

Im(z)

b

1

1

2

2

3

4

43–3–4

–4

–3

–2

–2

–1–1

Re(z)

Im(z)

z1 + z2

z1

z1 – z2

–z2

z2

0

z1 + z2 = (2 + i) + (−1 + 3i)

= 1 + 4i

z1 − z2 = (2 + i) − (−1 + 3i)

= 3 − 2i

Rotation about the originWhen the complex number 2 + 3i is multiplied by –1 the result is −2 − 3i .

This can be considered to be achieved through a rotation of 180◦ about the origin. When the

complex number 2 + 3i is multiplied by i,

i.e. i(2 + 3i) = 2i + 3i2

= 2i − 3

= −3 + 2i

the result can be seen to be achieved through

a rotation of 90◦in an anticlockwise direction

about the origin. If −3 + 2i is multiplied by

i the result is −2 − 3i . This is again achieved

through a rotation of 90◦ in an anticlockwise

direction about the origin.

Re(z)

Im(z)

0

–3 + 2i

–2 – 3i

2 + 3i

Example 18

If z1 = 1 − 4i and z2 = −2 + 2i , find z1 + z2 algebraically and illustrate z1 + z2 on an

Argand diagram.

SolutionIm(z)

Re(z)

z1

1 2 3 4–1

–2

–3

–4

–1–2–3–4

01

2

3

z1 + z2

z2

z1 + z2 = (1 − 4i) + (−2 + 2i)

= −1 − 2iSAMPLE


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Exercise 16D

1 Write down the complex numbers

represented on the following Argand diagram.

Example 16

5–1

–2

–1–2

1

2

Im(z)

Re(z)1 2 3 44

–3

–4

F

0

AE

D

C

B

–3–4–5

3

2 Represent each of the following complex numbers as points on an Argand plane.Example 17

a 3 − 4i b −4 + i c 4 + i d −3 + 0i e 0 − 2i f −5 − 2i

3 If z1 = 6 − 5i and z2 = −3 + 4i , find algebraically and represent on an Argand diagram.Example 18

a z1 + z2 b z1 − z2

4 If z = 1 + 3i , represent on an Argand diagram

a z b z c z2 d −z e1

z

5 If z = 2 − 5i , represent on an Argand diagram

a z b zi c zi2 d zi3 e zi4

16.5 Solving equations over the complex fieldQuadratic equations for which the discriminant is less than zero have no solutions for the real

numbers. The introduction of the complex number enables such quadratic equations to be

solved. Further solutions to higher degree polynomials may also be found using complex

numbers. Solution of higher degree polynomials appears in the Specialist Mathematics course.

In this chapter only quadratics will be considered.

Sum of two squaresEarlier it was seen that the product of a complex number a + bi and its conjugate a − bi

yielded the result

(a + bi)(a − bi) = a2 + b2

Hence sums of two squares can be factorised enabling equations of the form z2 + a2 = 0 to

be solved.SAMPLE


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Example 19

Solve the equations

a z2 + 16 = 0 b 2z2 + 6 = 0

Solution

a z2 + 16 = 0

∴ z2 − 16i2 = 0

(z + 4i)(z − 4i) = 0

∴ z = ±4i

b 2z2 + 6 = 0

∴ 2(z2 + 3) = 0

∴ 2(z2 − 3i2) = 0

∴ 2(z +√

3i)(z −√

3i) = 0

∴ z = ±√

3i

Solution of quadratic equationsTo solve quadratic equations where the discriminant is less than zero, still use the quadratic

formula in the usual way.

Example 20

Solve the equation 3z2 + 5z + 3 = 0

Solution

Using the quadratic formula

z = −5 ± √25 − 36

6

= −5 ± √−11

6

= 1

6(−5 ±

√11i)

Using the TI-NspireEach of the expressions in the above examples can be factorised using cFactor from the

Complex submenu of Algebra, for example,

cfactor (z2 + 16, z).

Each of the equations in the above examples can besolved using cSolve from the

Complex submenu of Algebra, for example

cSolve (3z2 + 5z + 3 = 0, z).

SAMPLE


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Using the Casio ClassPadTo factorise in the above examples, ensure the mode is set to Cplx.

Enter and highlight the expression z2 + 16 then

tap Interactive, Transformation, rFactor.

To solve in the above examples, the usual method for solving equations is used. For

example, enter and highlight 3z2 + 5z + 3 = 0 then tap Interactive,Equation/inequality, solve and ensure that the variable selected is z.

Exercise 16E

1 Solve each of the following equations over C.Examples 19, 20

a z2 + 4 = 0 b 2z2 + 18 = 0 c 3z2 = −15

d (z − 2)2 + 16 = 0 e (z + 1)2 = −49 f z2 − 2z + 3 = 0

g z2 + 3z + 3 = 0 h 2z2 + 5z + 4 = 0 i 3z2 = z − 2

j 2z = z2 + 5 k 2z2 − 6z = −10 l z2 − 6z = −14

16.6 Polar form of a complex numberEarlier in this chapter it was shown that points on a cartesian plane (x, y) may be represented in

terms of polar coordinates [r, �]. Similarly, complex numbers may be represented in polar

form.

Recalling that x = r cos � and y = r sin � where

r2 = x2 + y2 then the point P in the complex plane

corresponding to the complex number in cartesian form,

z = x + yi , may be represented as shown in the diagram.

z = r cos � + r sin � i

= r (cos � + sin i)

Im(z)

Re(z)

P(x + iy)

yr

xθ

0

The polar form is abbreviated to z = r cis �.

r =√

x2 + y2 is called the absolute value or modulus of z. It is denoted by mod z or |z|.Remember that � is measured in an anticlockwise direction from the horizontal axis.

Note: The same point may be represented a number of ways in polar form, since

cos � = cos(� ± 2n�) and sin � = sin (� ± 2n�), where n ∈ Z , the polar form of a complex

number is not unique.

i.e. z = r cis � = r cis (� + 2n�), n ∈ Z

SAMPLE


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Usually the interval −� < � ≤ � is used. The corresponding value of � is called the principal

value of the argument of z and is denoted by Arg z.

i.e. − � < Arg z ≤ �

Example 21

Express in polar form the following complex numbers

a z = 1 + √3i b z = 2 − 2i

Solution

a First note that z = 1 + √3 i is a point in the 1st quadrant.

∴ 0 < � <�

2Now x = 1 and y = √

3

Therefore r = √1 + 3

= 2

also �= �

3(since cos � = 1

2and sin � =

√3

2)

∴ z = 1 + √3i

= 2 cis�

3

b z = 2 − 2i is a point in the 4th quadrant.

∴ −�

2< � < 0

Now x = 2 and y = 2

Therefore r = √4 + 4

= √8

= 2√

2

Also � = −�

4(since cos � = 1√

2and sin � = −1√

2)

∴ z = 2 − 2i

= 2√

2 cis

(−�

4

)

SAMPLE


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Example 22

Express in cartesian form z = 2 cis

(−2�

3

)

Solution

x = r cos � = 2 cos

(−2�

3

)

= 2

(−1

2

)= −1

y = r sin � = 2 sin

(−2�

3

)

= 2

(−√

3

2

)

= −√

3

∴ z = 2 cis

(−2�

3

)

= −1 −√

3i

Multiplication and division in polar formIf z1 = r1 cis � and z2 = r2 cis �2

Then z1z2 = r1r2 cis (�1 + �2)

andz1

z2= r1

r2cis (�1 − �2)

These results may be proved using the addition formulas for sine and cosine established in

Chapter 11. This is left as an exercise for the reader.

Example 23

If z1 = 2 cis 30◦ and z2 = 4 cis 20◦ find the product z1z2 and represent it on an Argand

diagram.

SolutionIm(z)

Re(z)

30°50°

z2

z1z2

20°0

z1

z1z2 = r1r2 cis (�1 + �2)

= 2 × 4 cis (20◦ + 30◦)

= 8 cis 50◦

SAMPLE


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Example 24

If z1 = 3 cis�

2and z2 = 2 cis

(5�

6

), find the product z1z2.

Solution

z1z2 = r1r2 cis (�1 + �2)

= 6 cis

(�

2+ 5�

6

)

= 6 cis

(4�

3

)

∴ z1z2 = 6 cis

(−2�

3

)since −� < Arg z ≤ �

Example 25

If z1 = −√3 + i and z2 = 2

√3 + 2i , find the quotient

z1

z2and express it in cartesian form.

Solution

First express z1 and z1 in polar form.

|z1| = √3 + 1 Arg z1 = 5�

6, since sin �1 = 1

2and cos �1 = −√

3

2= 2 where Arg z1 = �1

|z2| = √12 + 4 Arg z2 = �

6, since sin �2 = 1

2and cos �2 =

√3

2= 4 where Arg z2 = �2

∴ z1 = 2 cis

(5�

6

)and z2 = 4 cis

�

6z1

z2= r1

r2cis (�1 − �2)

= 2

4cis

(5�

6− �

6

)

= 1

2cis

(2�

3

)

In cartesian formz1

z2= 1

2cos

(2�

3

)+ 1

2sin

(2�

3

)i

= 1

2

(−1

2

)+ 1

2

(√3

2

)i

= −1

4(1 −

√3)i

SAMPLE


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Exercise 16F

1 Express each of the following in the simplest polar form.Example 21

a 1 + √3i b 1 − i c −2

√3 + 2i

d −4 − 4i e 12 − 12√

3i f −1

2+ 1

2i

2 Express each of the following in the form x + yi .Example 22

a 3 cis�

2b

√2 cis

�

3c 2 cis

�

6d 5 cis

3�

4

e 12 cis5�

6f 3

√2 cis

−�

4g 5 cis

4�

3h 5 cis

−2�

3

3 Simplify the following and express the answers in cartesian form.Examples 23, 24, 25

a(

2 cis�

6

).(

3 cis�

12

)b

(4 cis

�

12

).(

3 cis�

4

)

c(

cis�

4

).

(5 cis

5�

12

)d

(12 cis

−�

3

).

(3 cis

2�

3

)

e

(12 cis

5�

6

).(

3 cis�

2

)f (

√2 cis �).

(√3 cis

−3�

4

)

g

(10 cis

�

4

)(

5 cis�

12

) h

(12 cis

−�

3

)

3 cis2�

3

i

(12

√8 cis

3�

4

)(

3√

2 cis�

12

) j

(20 cis

−�

6

)(

8 cis5�

6

)

SAMPLE


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Review


Chapter summary

The polar coordinates [r, �] may be represented as follows.P[r, θ]

θ

polepolar axisO

� is measured in an anticlockwise direction

from the polar axis.

For conversion of coordinates from cartesian to polar and vice versa

x = r cos �, y = r sin � and hence x2 + y2 = r2.

Therefore[2,

�

3

]describes the same point as

(2 cos

�

3, 2 sin

�

3

)= (1,

√3)

For (1, −1), r = √2

1 = √2 cos � and − 1 = √

2 sin �

Therefore cos � = 1√2

and sin � = −1√2

and � = −�

4

(1, −1) =[√

2,−�

4

]

For conversion of an equation from cartesian to polar use

x = r cos �, y = r sin � and x2 + y2 = r2

Therefore x + y = 1

becomesr cos � + r sin � = 1

i.e. r (cos � + sin �) = 1

Consider y = x2

This becomes

r sin � = r2 cos2 �

∴ r sin � − r2 cos2 � = 0

∴ r (sin � − r cos2 �) = 0

r = 0 or sin � = r cos2 �

r = 0 is the pole. The second equation becomes r = tan �

cos �

i is an imaginary number with the property i2 = −1.

C, the set of complex numbers, is defined by C = {a + bi : a, b ∈ R}.Real numbers and Imaginary numbers are subsets of C.

Re (z) is the real component of z.

Im (z) is the value of the imaginary component of z.

SAMPLE


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Rev

iew


z1 = z2 ⇔ Re (z1) = Re (z2) and Im (z1) = Im (z2)

The Argand diagram is a geometrical representation of C.

Let z1 = a + bi and z2 = c + di, then z1z2 = (ac − bd) + (ad + bc)i

The modulus of z, |z|, is the distance from the origin of the point represented by z.

The argument of z, arg z, is an angle measured anticlockwise about the origin from the

positive direction of the x axis to the line joining the origin to z.

The Argument of z, Arg z, is arg z expressed as an angle in the interval (−�, �].

The modulus-argument form of the complex number z is given as:

z = r (cos � + i sin �) where r = |z|cos � = Re (z)

|z| and sin � = Im (z)

|z|r (cos � + i sin �) is usually written as r cis �

The complex conjugate of z is denoted by z, where z = Re (z) − Im (z)i ; zz, (z + z) ∈ R

The division of complex numbers:z1

z2= z1z2

|z2|2Multiplication and division of the modulus-argument form

Let z1 = r1 cis �1, z2 = r2 cis �2

Then z1z2 = r1r2 cis (�1 + �2),z1

z2= r1

r2cis (�1 − �2)

Multiple-choice questions

1 The polar coordinates [−3, 30◦] define a point that can also be described by

A [3, 30◦] B [3, −30◦] C [3, 150◦] D [3, −150◦] E [−3, 150◦]

2 The polar coordinates of point A are

A [2, 40◦] B [2, −40◦] C [2, 140◦]

D [4, 40◦] E [2, −140◦]

A

40°2

ZO3 The polar coordinates of the point with cartesian coordinates (−1, −√

3) are

A

[2,

−4�

3

]B

[−2,

�

3

]C

[2,

�

3

]D

[2,

2�

3

]E

[−2, −�

3

]

4 The cartesian coordinates of the point with polar coordinates[3,

�

6

]are

A (3,√

3) B

(3,

1√3

)C

(3

2,

3√

3

2

)D

(3√

3

2,

3

2

)E

(3

2,

√3

2

)

5 The polar equation of the circle with centre[3,

�

2

]and radius 3 is

A r = 3 sin � B r = √3 C r = 3 D r = 3 cos � E r = 6 sin �

SAMPLE


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Review


6 The graph of r cos � = 2 is

A

O

B

O

C

O

D

O

E

O

7 The polar equation of the circle with cartesian equation x2 + y2 = 16 is

A r = 16 B r = 4 sin � C r2 + cos2 � = 4 D r = 4 cos � E r = 4

8 If u = 1 + i , then1

2 − u=

A −1

2− 1

2i B

1

5+ 2

5i C

1

2+ 1

2i D −1

2+ 1

5i E 1 + 5i

9 The point C on the Argand diagram represents the complex number z. Which point

represents the complex number i × z?

A A B B C C D D E E

Re(z)

B

D E

AC

Im(z)

10 If |z| = 5 then

∣∣∣∣1

z

∣∣∣∣ =

A1√5

B − 1√5

C1

5D −1

5E

√5

Short-answer questions (technology-free)

1 Graph each of the following.

a [3, �] b[2,

�

3

]c [−2, 210◦] d

[−3,

11�

6

]

2 Find the cartesian coordinates of the points in 1.

SAMPLE


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Rev

iew


3 Graph each of the following.

a {[r, �] : r = 3} b{

[r, �] : � = �

3

}c {[r, �] : r = −4} d

{[r, �] : � = −5�

4

}4 Express each of the following in polar form.

a (3, 3) b

(√3

2,−1

2

)c

(−5

2,−5

√3

2

)d (4

√2, −4

√2)

5 Transform each of the following equations from cartesian to polar form.

a x2 + y2 = 16 b x2 + y2 = 9 c y2 = 8x

d x2 = 4y e x2 + 4y2 = 64 f 2x − y + 2 = 0

6 Transform each of the following equations from polar to cartesian form.

a r = 5 b r = 3 sin � c r2 cos 2� = 9

d r (1 − 2 cos �) = 8 e r (2 − cos �) = 7 f r (1 − sin �) = 1

7 For z1 = m + in and z2 = p + iq , express each of the following in the form a + ib.

a 2z1 + 3z2 b z2 c z1z2

dz1

z2e z1 + z1 f (z1 + z2)(z1 − z2)

g1

z1h

z2

z1i

3z1

z2

8 In the following, z = 1 − √3i . Express each in the form a + ib and mark each of the

following on an Argand diagram.

a z b z2 c z3 d1

ze z f

1

z9 Write each of the following in polar form.

a 1 + i b 1 − √3i c 2

√3 + i

d 3√

2 + 3√

2i e −3√

2 − 3√

2i f√

3 − i

10 Write each of the following in cartesian form.

a −2 cis�

3b 3 cis

�

4c 3 cis

3�

4

d −3 cis

(−3�

4

)e 3 cis

(−5�

6

)f

√2 cis

(−�

4

)

Extended-response questions

1 a Plot the graphs of r = 2 sin � and r = 2 cos �.

b Write the corresponding cartesian equation for each of these relations.

c Describe the curves you obtain from the polar equations r = 2a sin � or r = 2a cos �

where a is a non-zero constant.SAM

PLE


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Review


2 Investigate each of the families of graphs defined by:

a r = a + b sin � or r = a + b cos � where a and b are non-zero constants

b r2 = a2 sin 2�, r2 = −a2 sin 2�, r2 = a2 cos 2�, r2 = −a2 cos 2� where a is a positive

constant

c r = a�

d r = a sin n� and r = a cos n� where a is a non-zero constant

3 a Find the exact solutions in C for the equation z2 − 2√

3z + 4 = 0.

b i Plot the two solutions from a on an Argand diagram.

ii Find the equation of the circle, with centre the origin, which passes through these

two points.

iii Find the value of a ∈ Z such that the circle passes through (0, ±a)

4 Let z be a complex number with |z| = 6. Let A be the point representing z. Let B be the

point representing (1 + i)z.

Find

a |(1 + i)z| b |(1 + i)z − z|c Prove that OAB is an isosceles right-angled triangle.

5 Let z = 1√2

+ 1√2

i

a On an Argand diagram O, A, Z , P, Q represent the complex numbers 0, 1, z, 1 + z and

1 − z respectively. Show these points on a diagram.

b Prove that the magnitude of ∠POQ = �

2. Find the ratio

|OP||OQ| .

SAMPLE


Documents

P1: FXS/ABE P2: FXS CHAPTER 16 - Cambridge University Press · P1: FXS/ABE P2: FXS 9780521740494c16a.xml CUAU033-EVANS September 12, 2008 10:50 414 Essential Advanced General Mathematics