Number systems: complex numbersmathsbooks.net/Maths Quest 11C for Queesnalnd/ch 02... · 1. The...

In thisIn this chachapterpter2A Introduction to complex

numbers2B Basic operations using

complex numbers2C Conjugates and division

of complex numbers2D Radians and coterminal

angles2E Complex numbers in polar

form2F Basic operations on complex

numbers in polar form

syllabussyllabusrrefefererenceenceCore topic:Real and complex number systems

2

Number systems:complexnumbers

MQ Maths C Yr 11 - 02 Page 77 Tuesday, October 9, 2001 3:59 PM

78

M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Introduction to complex numbers

In 1545, the Italian mathematician Girolamo Cardano proposed (what was then) a star-tling mathematical expression:

This was a valid expression, yet it included the square root of a negative number,which seemed ‘impossible’.

What is a number such as or and how does it relate to a real number, andwhat does it signify in mathematics?

Back in chapter 1, you may recall that our definition of real numbers includedwhole numbers, fractions, irrational and rational numbers as subsets of the realnumber set. Whenever the square root of a negative number was encountered wasthis classified as real? Where did we sometimes encounter such numbers in calcu-lations? Solution of quadratic equations sometimes brought these numbers to theforeground. What was the difference between these two situations:

x

2

+

3

x

−

6

=

0and

x

2

+

3

x

+

1

=

0? How did the solutions to these equations relate to propertiesof the associated parabolas?

In terms of the mathematics that you have studied so far, these square roots of nega-tive numbers have some significance.

But why did the square roots of negative numbers become central to the study of anew set of numbers called the complex numbers? It was partly curiosity and partlybecause mathematicians such as Diophantus (the Greek mathematician) and Leibniz(the German mathematician) found that real numbers could not solve all equations.Eventually scientists and engineers discovered their uses. Complex numbers are nowused extensively in the fields of physics and engineering in areas such as electric cir-cuits and electromagnetic waves. Combined with calculus theory, complex numbersform an important part of the study of mathematics known as

complex analysis

.

Square root of a negative number

The quadratic equation

x

2

+

1

=

0 has no solutions for

x

in the Real Number System

R

because the equation yields and there is no real number which, whensquared, gives

−

1 as the result. If, however, we define an ‘imaginary number’ denoted

by

i

such that

i

=

, then becomes

x

=

±

i

. For the general case

x

2

+

a

2

=

0, with

a

∈

R

, we can write:

x

=

=

=

±

(

×

)

=

±

ai

Powers of

i

will produce

±

i

or

±

1. We have

i

2

= −

1,

i

3

=

i

2

×

i

= −

1

×

i

= −

i

,

i

4

=

i

2

×

i

2

= −

1

× −

1

=

1,

i

6

=

(

i

2

)

3

=

(

−

1)

3

= −

1 and so on. The pattern is quiteobviously that even powers of

i

result in 1 or

−

1 and odd powers of

i

result in

i

or

−

i

.

Definition of a complex numberA complex number (generally denoted by the letter z) is defined as a quantityconsisting of a real number added to a multiple of the imaginary unit i. For realnumbers x and y, x + yi is a complex number. This is referred to as the standard orCartesian form.

40 (5 15–+ )(5 15– )–=

1– 15–

x 1–±=

1– x 1–±=

a2–±

1 a2×–±

1– a2


C h a p t e r 2 N u m b e r s y s t e m s : c o m p l e x n u m b e r s

79

C

=

{

z

:

z

=

x

+

yi

where

x

,

y

∈

R

}

defines the set of complex numbers.The real part of

z

is

x

and is written as Re (

z

). That is, Re (

z

)

=

x

.The imaginary part of

z

is

y

and is written as Im (

z

). That is, Im (

z

)

=

y

.

Note

: Every real number

x

can be written as

x

+

0

i

and so the set of real numbers is asubset of the set of complex numbers. That is,

R

⊂

C

.

Using the imaginary number i, write a simplified expression for:a b .

THINK WRITE

a Express the square root of −16 as the product of the square root of 16 and the square root of −1.

a =

Substitute i for . = × iTake the square root of 16. = 4i

b Express the square root of −5 as the product of the square root of 5 and the square root of −1.

b =

Substitute i for . = × i

=

16– 5–

1 16– 16 1–×

2 1– 16

3

1 5– 5 1–×

2 1– 5

i 5

1WORKEDExample

Write down the real and imaginary parts of the following complex numbers, z.a z = −3 + 2i b z = − i

THINK WRITE

a The real part is the ‘non-i’ term. a Re (z) = −3

The imaginary part is the coefficient of the i term.

Im (z) = 2

b The real part is the ‘non-i’ term. b Re (z) = 0

The imaginary part is the coefficient of the i term.

Im (z) = −

12---

1

2

1

212---

2WORKEDExample

Write i8 + i5 in the form x + yi where x and y are real numbers.

THINK WRITE

Simplify both i8 and i5 using the lowest possible power of i.

i8 = (i2)4 = (−1)4 = 1i5 = i4 × i = (i2)2 × i = (−1)2 × i = 1 × i = i

Add the two answers. i8 + i5 = 1 + i

1

2

3WORKEDExample

MQ Maths C Yr 11 - 02 Page 79 Wednesday, November 5, 2003 10:26 AM

80 M a t h s Q u e s t M a t h s C Ye a r 1 1 f o r Q u e e n s l a n d

Simplify z = i4 − 2i2 + 1 and w = i6 − 3i4 + 3i2 − 1 and show that z + w = −4.

THINK WRITE

Replace terms with the lowest possible powers of i (remember i2 = −1).

i4 − 2i2 + 1 = (i2)2 − 2 × −1 + 1= (−1)2 + 2 + 1= 4

i6 − 3i4 + 3i2 − 1 = (i2)3 − 3(i2)2 + 3 × −1 − 1= (−1)3 − 3(−1)2 − 3 − 1= −1 − 3 − 3 − 1= −8

Add the two answers. z + w = i4 − 2i2 + 1 + i6 − 3i4 + 3i2 − 1z + w = 4 − 8z + w = −4

1

2

4WORKEDExample

Evaluate each of the following.

a Re (7 + 6i) b Im (10) c Re (2 + i − 3i3) d Im

THINK WRITE

a The real part of the complex number7 + 6i is 7.

a Re (7 + 6i) = 7

b The number 10 can be expressed in complex form as 10 + 0i and so the imaginary part is 0.

b Im (10) = Im (10 + 0i)= 0

c Simplify 2 + i − 3i3. c Re (2 + i − 3i3) = Re (2 + i − 3i × i2)= Re (2 + i + 3i)= Re (2 + 4i)

The real part is 2. = 2

d Simplify the numerator of

.

d Im = Im

= Im

Simplify by dividing the numerator by 2.= Im

= Im (1 − i)The imaginary part is −1. = −1

1 3i– i2– i3–2

----------------------------------

1

2

1

1 3i– i2– i3–2

----------------------------------

1 3i– i2– i3–2

---------------------------------- 1 3i– 1 i+ +

2-------------------------------

2 2i–2

--------------

2 2 1 i–( )2

------------------

3

5WORKEDExample


C h a p t e r 2 N u m b e r s y s t e m s : c o m p l e x n u m b e r s 81

Introduction tocomplex numbers

1 Using the imaginary number i, write down expressions for:

2 Write down the real and imaginary parts, respectively, of the following complexnumbers, z.

3 Write each of the following in the form x + yi, where x and y are real numbers.

4 Simplify z = i6 + 3i7 − 2i10 − 3 and w = 4i8 − 3i11 + 3 and show that z + w = 5.

5 Evaluate each of the following.

6 Write in the form x + yi, where x and y are real numbers.

7

a The value of Re (i + i3 + i5) is:

b The value of Im [i(2i4 − 3i2 + 5i)] is:

c The expression i + i2 − i3 + i4 − i5 + i6 simplifies to:

d If which one of the statements below is true?

8 If n is an even natural number show that .

a b c d

e f g h

a 9 + 5i b 5 − 4i c −3 − 8i d 11i − 6e 27 f 2i g –5 + i h –17i

a i9 + i10 b i9 − i10 c i12 + i15 d i7 − i11

e i5 + i6 − i7 f i(i13 + i16) g 2i − i2 + 2i3 h 3i + i4 – 5i5

a Re (−5 + 4i) b Re (15 − 8i) c Re (12i)d Im (1 − 6i) e Im (3 + 2i) f Im (8)

g Re (i5 − 3i4 + 6i6) h Im

A 2 B −1 C 3 D 1 E 0

A 0 B −5 C 5 D 10 E 4

A i B 0 C i − 1 D i + 2 E −i

A f(i) = 2 + i B Re [f(i)] = 5 C Im [f(i)] = −D f(i) = 1 − i E f(i) = 0

remember1. The ‘imaginary number’ i has the property that i2 = −1.2. A complex number z is of the form z = x + yi where x, y ∈R.3. The real part of z is x and is written as Re (z).4. The imaginary part of z is y and is written as Im (z).

remember

2AWORKEDExample

1

Mathcad

Complex 19– 25– 49– 3–

11– 7– 49---– 36

25------–

WORKEDExample

2

WORKEDExample

3

WORKEDExample

4

WORKEDExample

5

4i9 5i14– 2i7–3

------------------------------------

3 i3 i– 2+i2 i4–

---------------------–

mmultiple choiceultiple choice

f i( ) 1 i i2 … i11+ + + +4

----------------------------------------------=14---

1–( )n2---

in=



Basic operations using complex numbers

Complex numbers can be added, subtracted, multiplied and divided. In general, thesolutions obtained when performing these operations are presented in the standard formz = x + yi.

Argand diagramsWe know that an ordered pair of real numbers (x, y) can be represented on the Cartesianplane. Similarly, if we regard the complex number x + yi as consisting of the orderedpair of real numbers (x, y), then the complex number z = x + yi can be plotted as a point(x, y) on the complex number plane.

Complex numbers in quadratic equations

In your junior mathematics studies you graphed quadratics and found the real roots of the expressions using the formula for the solution of a general quadratic equation of the form ax2 + bx + c. Sometimes the values for a, b and c meant that the value under the radical sign was negative; that is, it had a negative discriminant (for example, ). You might have been told that this meant there were ‘no real roots’ for this quadratic. That was correct, but only half the answer. Follow the steps below and you will hopefully develop a better understanding of the results you obtain.

The following formulas are included for your assistance:

x = , turning point x =

Step 1 Use the formula for the solution of a quadratic equation to find the roots of:

y = x2 − 2x + 3 ....................(A)Interpret this result.

Step 2 Use the formula for the x-coordinate of the turning point and substitute this into the original quadratic to find the y-coordinate of this turning point.

Step 3 Graph this quadratic equation using the information from steps 1 and 2.

Repeat steps 1–3 with the following quadratic equations. Note the effect of the negative discriminant in equation (C).

y = x2 − 2x + 1 ..................................(B)y = x2 − 2x − 2 ..................................(C)

Graphically, we can see that there are no real values of x that satisfy the equation x2 − 2x − 2 = 0.

16–

–b b2 4ac–±

2a------------------------------------- b–

2a------


C h a p t e r 2 N u m b e r s y s t e m s : c o m p l e x n u m b e r s 83This is also referred to as the Argand plane or an Argand diagram in recognition of

the work done in this area by the Swiss mathematician Jean-Robert Argand.The horizontal axis is referred to as the Real axis and the vertical axis is referred to

as the Imaginary axis.The points A, B, C, D and E shown on the

Argand diagram at right represent the complexnumbers 3 + 0i, 0 + 2i, −4 + 5i, −3 − 4i and 2 − 2irespectively.

This method of representation is a useful way ofillustrating the properties of complex numbersunder the operations of addition, subtraction andmultiplication.

Addition of complex numbersAddition is performed by adding the real and imaginary parts separately.

If z = a + bi and w = c + di then z + w = (a + c) + (b + d)i where Re (z + w) = Re (z) + Re (w) and Im (z + w) = Im (z) + Im (w).

Geometric representationIf z1 = x1 + y1i and z2 = x2 + y2i then z2 + z1 = (x2 + x1) + (y2 + y1)i. If a directed linesegment connects the origin (0 + 0i) to each of the points z1, z2 and z1 + z2, then theaddition of two complex numbers can be associated with standard methods of additionof the directed line segments.

The figure at right illustrates the situation forz2 + z1, with, say, positive values for x1, x2, y1, y2and x1 < x2 and y1 < y2.

Note: The origin, z1, z2 and z2 + z1 form a parallelogram.You will use this concept later in this course when youstudy vector addition.

Subtraction of complex numbersIf we write z − w as z + −w we can use the rule for addition of complex numbers toobtain z + −w = (a + bi) + − (c + di)

= a + bi − c − di= (a − c) + (b − d)i

If z = a + bi and w = c + di then z − w = (a − c) + (b − d)i.

Geometric representationIf z1 = x1 + y1i and z2 = x2 + y2i then z2 − z1 = (x2 − x1)+ (y2 − y1)i. If a directed line segment connects theorigin (0 + 0i) to each of the points z1, z2 and z2 − z1then the subtraction of two complex numbers can alsobe associated with standard methods of the addition ofdirected line segments. The figure at right illustratesthe situation for z2 − z1, again with positive values forx1, x2, y1, y2 and x1 < x2 and y1 < y2.

–4 –3 –2–1 4321

–4–3–2–1

12345

Im (z) (Imaginary axis)

(Real axis)Re (z)

C

D

E

A

B

x1 x2

(y1 + y2)

(x1 + x2)

y2y1

z1 z2

z1 + z2

0

Im (z)

Re (z)

x1 x2

(y2 – y1)

(x2 – x1)

y2

y1

z1 z2 – z1

z2

0

Im (z)

Re (z)



Multiplication by a constant (or scalar)If z = x + yi and k∈Rthen kz = k(x + yi)

= kx + kyiFor k > 1, the product kz can be illustrated as shown at right.

The ratio of corresponding sides of the two triangles is k:1.A similar situation exists for k < 1.So when a complex number is multiplied by a constant,

this produces a directed line segment in the same direction (or at 180 degrees if k < 0)which is larger in length if k > 1 or smaller if 0 < k < 1. Geometrically this is called atransformation or dilation, which means magnifying or decreasing by a constant factor.

For z = 8 + 7i, w = −12 + 5i and u = 1 + 2i, calculate:a z + w b w − z c u − w + z.THINK WRITEa Use the addition rule for complex numbers. a z + w = (8 + 7i) + (−12 + 5i)

= (8 − 12) + (7 + 5)i= −4 + 12i

b Use the subtraction rule for complex numbers.

b w − z = (−12 + 5i) − (8 + 7i)= (−12 − 8) + (5 − 7)i= −20 − 2i

c Use both the addition rule and the subtraction rule.

c u − w + z = (1 + 2i) − (−12 + 5i) + (8 + 7i)= (1 + 12 + 8) + (2 − 5 + 7)i= 21 + 4i

6WORKEDExample

x kx

z

kzky

y

0

Im (z)

Re (z)

If z = 3 + 5i, w = 4 − 2i and v = 6 + 10i, evaluate:a 3z + w b 2z − v c 4z − 3w + 2v.THINK WRITEa Calculate 3z + w by substituting values

for z and w.a 3z + w = 3(3 + 5i) + (4 − 2i)

= (9 + 15i) + (4 − 2i)Use the rule for adding complex numbers. = (9 + 4) + (15 − 2)i

= 13 + 13ib Calculate 2z − v by substituting values for

z and v.b 2z − v = 2(3 + 5i) − (6 + 10i)

Use the rule for subtraction of complex numbers. = 6 + 10i − 6 − 10i= 0 + 0i= 0

c Calculate 4z − 3w + 2v by substituting values for z, w and v.

c 4z − 3w + 2v= 4(3 + 5i) − 3(4 − 2i) + 2(6 + 10i)

Use the addition rule and the subtraction rule to simplify.

= 12 + 20i − 12 + 6i + 12 + 20i= 12 + 46i

1

2

1

2

1

2

7WORKEDExample

MQ Maths C Yr 11 - 02 4 Tuesday, October 9, 2001 3:59 PM


Multiplication of two complex numbersSo far we have shown that complex numbers can be plotted on an Argand diagram;adding and subtracting them is geometrically equivalent to adding and subtractingdirected line segments and multiplication by a positive constant is equivalent toextending or shrinking the directed line segment without altering the direction.

What geometrical interpretation, if any, can be given to multiplication of two (ormore) complex numbers?

The multiplication of two complex numbers also results in a complex number.If z = a + bi and w = c + di

then z × w = (a + bi)(c + di)= ac + adi + bci + bdi2

= (ac − bd) + (ad + bc)i (since i2 = −1)

If z = a + bi and w = c + di then z × w = (ac − bd) + (ad + bc)i.

If z = 6 − 2i and w = 3 + 4i express zw in standard form.THINK WRITE

Expand the brackets. zw = (6 − 2i)(3 + 4i)= 18 + 24i − 6i − 8i2

Express in the form x + yi by substituting −1 for i2 and simplifying the expression using the addition and subtraction rules.

= 18 + 24i − 6i + 8= 26 + 18i

1

2

8WORKEDExample

Simplify (2 − 3i)(2 + 3i).THINK WRITE

Expand the brackets. (2 − 3i)(2 + 3i) = 4 + 6i − 6i − 9i2

Substitute −1 for i2 and simplify the expression. = 4 − 9 × −1= 13

12

9WORKEDExample

Determine Re (z2w) + Im (zw2) for z = 4 + i and w = 3 − i.

Continued over page

THINK WRITEExpress z2w in the form x + yi. z2w = (4 + i)2(3 − i)

= (16 + 8i + i2)(3 − i)= (16 + 8i − 1)(3 − i)= (15 + 8i)(3 − i)= 45 − 15i + 24i − 8i2

= 53 + 9iThe real part, Re (z2w) is 53. Re (z2w) = 53

1

2

10WORKEDExample


86


Equality of two complex numbers

If

z

=

a

+

bi

and

w

=

c

+

di

then

z

=

w

if and only if

a

=

c

and

b

=

d

.

The condition ‘if and only if’ (sometimes written in short form as iff ) means that bothof the following situations must apply.1. If

z

=

w

then

a

=

c

and

b

=

d

.2. If

a

=

c

and

b

=

d

then

z

=

w

.

Plotting complex numbers

You will need 1 cm square grid paper, a ruler and a protractor.

For

z

= −

3

+

2

i

and

w

=

5

+

i

1

Accurately plot

z

and

w

on an Argand diagram.

2

Find

zw

and plot this on the same diagram.

3

Measure each length and angle from the positive end on the Real axis. Do you notice any pattern between the numbers you started with and your result?

4

Try this with some other complex numbers. Plot your original complex numbers accurately and plot the product. Test your original hypothesis.

THINK WRITE

Express zw2 in the form x + yi. zw2 = (4 + i)(3 − i)2

= (4 + i)(9 − 6i + i2)= (4 + i)(8 − 6i)= 32 − 24i + 8i − 6i2

= 38 − 16i

The imaginary part, Im (zw2) is −16. Im (zw2) = − 16

Calculate the value of Re (z2w) + Im (zw2).

Re (z2w) + Im (zw2) = 53 − 16= 37

3

4

5

Find the values of x and y that satisfy (3 + 4i)(x + yi) = 29 + 22i.

THINK WRITE

Write the left-hand side of the equation LHS = (3 + 4i)(x + yi) Expand the left-hand side of the equation. LHS = 3x + 3yi + 4xi + 4yi2

Express the left-hand side in the form a + bi. LHS = (3x − 4y) + (4x + 3y)i

1

2

3

11WORKEDExample



Multiplication by iLet us examine the effect on z = x + yi after multiplication by i, i2, i3 and i4.

z = x + yiiz = i(x + yi) = −y + xi

i2z = −1z = −x − yi = −zi3z = i(i2z) = y − xi = −izi4z = i(i3z) = x + yi = z

The five points are shown on the complex plane at right.

It is observed that multiplying z by in, n ∈ N produces an anticlockwise rotation of 90n degrees.

THINK WRITE

Equate the real parts and imaginary parts to create a pair of simultaneous equations.

3x − 4y = 29 [1]4x + 3y = 22 [2]

Simultaneously solve [1] and [2] for x and y. 9x − 12y = 87 [3]

Multiply equation [1] by 3 and equation [2] by 4 so that y can be eliminated.

16x + 12y = 88 [4]

Add the two new equations and solve for x. Adding equations [3] and [4]:25x = 175

x = 7

Substitute x = 7 into equation [1] and solve for y.

Substitute x = 7 into equation [1]:3(7) − 4y = 29

21 − 4y = 29−4y = 8

y = −2

State the solution. Therefore x = 7 and y = −2.

Check the solution by substituting these values into equation [2].

Check: 4 × 7 + 3 × −2 = 22.

4

5

6

7

8

9

xiz

z or i 4z

i 3z

i 2z

x–x –y–y

Im (z)

Re (z)

rememberIf z = a + bi and w = c + di for a, b, c, d∈R then:1. z + w = (a + c) + (b + d)i2. z − w = (a − c) + (b − d)i3. kz = ka + kbi, for k∈R4. z × w = (ac − bd) + (ad + bc)i5. z = w if and only if a = c and b = d.

(Note: ‘If and only if’ can be written as iff.)

remember



Basic operations usingcomplex numbers

1 Represent each of the following complex numbers on an Argand diagram.

2 For z = 5 + 3i, w = −1 − 4i, u = 6 − 11i and v = 2i − 3 calculate:

3 If z = −3 + 2i, w = −4 + i and u = −8 − 5i, evaluate:

4 Using z, w, u and v from question 2 express each of the following in the form x + yi.

5 Simplify the following.

6 For z = −1 − 3i and w = 2 − 5i calculate z2w.

7 Determine Re (z2) − Im (zw) for z = 1 + i and w = 4 − i.

8 For z = 3 + 5i, w = 2 − 3i and u = 1 − 4i determine:

9 Find the values of x and y that satisfy each of the following.

10

If z = 8 − 7i and w = 3 + 4i, then:a Re (zw) is equal to:

b Im (w2) + Re (z2) is equal to:

c 3z − 2w is equal to:

11 If z = 2 + i and w = 4 − 3i then represent each of the following on an Argand diagram.

12 If z = 3 + 2i represent each of the following on the same Argand diagram.z, iz, i2z, i3z, i4z, i5z, −iz, −i2z

a 3 + i b 4 − 5i c −2 − 6id 3i + 7 e f

a z + w b u − z c w + vd u − v e w − z − u f v + w − z

a 3w b 2u + z c 4z − 3ud 3z + u + 2w e 2z − 7w + 9u f 3(z + 2u) − 4w

a zw b uv c wud zu e u2 f u(wv)

a (10 + 7i)(9 − 3i) b (3 − 4i)(5 + 4i) c (8 − 2i)(4 − 5i)d (5 + 6i)(5 − 6i) e (2i − 7)(2i + 7) f (9 − 7i)2

a Im (u2) b Re (w2)c Re (uw) + Im (zw) d Re (zu) − Im (w2)e Re (z2) − Re (zw) − Im (uz) f Re (u2w) + Im (zw2)

a (2 + 3i)(x + yi) = 16 + 11i b (5 − 4i)(x + yi) = 1 − 4ic (3i − 8)(x + yi) = −23 − 37i d (7 + 6i)(x + yi) = 4 − 33i

A −4 B 4 C 5 D 11 E 52

A 76 B 39 C 105 D 56 E −32

A 30 − 13i B 30 − 29i C 18 − 29i D 24 − 13i E 18

a z2 b zw c z + w d w − ze 3z + w f 2w − 4z g (z + w)2 h (w − z)3

2B

Mathca

d

Complex 2 5 2i– 8– i 3+

WORKEDExample

6

WORKEDExample

7

SkillSH

EET 2.1 WORKEDExample

8

WORKEDExample

9

WORKEDExample

10

WORKEDExample

11




Conjugates and division of complex numbers

The conjugate of a complex numberThe conjugate of a complex number is obtained by changing the sign of the imaginarycomponent.

If z = x + yi the conjugate of z is defined as = x − yi.

Conjugates are useful because the multiplication (or addition) of a complex numberand its conjugate results in a real number.Multiplication: zz– = (x + yi)(x − yi),

= x2 + y2

where x, y ∈ R, and x − yi and x + yi are conjugates. You will use this result whendividing complex numbers.Note: Compare this expression with the formula for the difference of two squares

(a − b)(a + b) = a2 − b2.

Addition: z + z– = x + yi + x − yi= 2x

Graphing complex conjugatesAs seen earlier, z and its conjugate can be written as

z = x + yi and z– = x − yiThe geometrical representation of z and z– is shown at right.Notice that the conjugate z– appears as a reflection of z in the Real axis.Other properties of conjugates include:1. z–

– = z

2. z1 ± z2 = z–1 ± z–2

3. z1z2 = z–1z–

2

4. = where z2 ≠ 0

z z

x

z = x + yi

z = x – yi

y

y

Re (z)

Im (z)

z1

z2-------

z1

z2--------

Write the conjugate of each of the following complex numbers.

a 8 + 5i b −2 − 3i c

THINK WRITE

a Change the sign of the imaginary component. a 8 − 5ib Change the sign of the imaginary component. b −2 + 3i

c Change the sign of the imaginary component. c

4 i 5+

4 i 5–

12WORKEDExample



Multiplicative inverse of a complex numberGiven a non-zero complex number z, there exists a complex number w such that

zw = 1, with w being the multiplicative inverse of z denoted by .

This example shows that if z = a + bi then .

Division of complex numbersThe application of conjugates to division of complex numbers will now be investigated.

Consider the complex numbers z = a + bi and w = c + di. To find in the form x + yi

we must multiply both the numerator and denominator by the conjugate of w to makethe denominator a real number only.(You might need to review rationalisation of the denominator which was discussed inchapter 1.)

If z = 5 − 2i and w = 7 − i show that .

THINK WRITE

Add the conjugates z– and . z–

Add z to w.

Write down the conjugate of z + w.

The conjugate of z + w equals z– . = z–

z w+ z w+=

1 w w+ 5 2i+( ) 7 i+( )+ 12 3i+= =2 z w+ 5 2i–( ) 7 i–( )+ 12 3i–= =

3 z w+ 12 3i+=

4 z w+ z w+ w+

13WORKEDExample

w z 1– 1z---= =

If z = 3 + 4i, determine z−1.THINK WRITE

Write z–1 as a rational expression:z−1 =

Multiply both the numerator and denominator by the conjugate of 3 + 4i.

=

=

Write the expression in the form x + yi. =

1

z 1– 1z---=

1z--- 1

3 4i+--------------=

21

(3 4i)+------------------- (3 4i)–

(3 4i)–------------------×

3 4i–25

--------------

33

25------ 4i

25------–

14WORKEDExample

z 1– a bi–a2 b2+-----------------=

zw----



=

= Multiply by the conjugate of c + di.

= Simplify the expressions in the numerator and in the denominator.

= Express in the form x + yi.

Thus we can state:

If z and w are complex numbers in the form x + yi, then can also be expressed in the form x + yi by simplifying:

¥

zw---- a bi+

c di+--------------

a bi+c di+-------------- c di–

c di–-------------×

ac bd+( ) bc ad–( )i+c2 d2+

-------------------------------------------------------

ac bd+c2 d2+------------------ bc ad–( )i

c2 d2+-------------------------+

zw----

zw---- w

w----

Express in standard form.

THINK WRITE

Multiply both the numerator and denominator by the conjugate of 2 − i to make the denominator real.

=

Expand the expressions obtained in the numerator and denominator.

=

Substitute −1 for i2 and simplify the expression.

=

=

=

2 i+2 i–-----------

12 i+2 i–---------- (2 i)+

(2 i)–--------------- (2 i)+

(2 i)+---------------×

24 4i i2+ +

4 i2–------------------------

34 4i 1–+

4 1+-----------------------

3 4i+5

--------------

35--- 4i

5-----+

15WORKEDExample

Complex numbers can be used to generate fractal patterns such as the ‘Julia Set’ shown.



If z = 3 + i, and determine Im (4z − w).

THINK WRITE

Substitute for z and w in 4z − w.Express 4z − w with a common denominator.

4z − w =

=

=

=

Remove i from the denominator by multiplying the numerator and denominator by the conjugate of 4 − i.

=

=

Simplify the expression so that it is in the form x + yi.

=

State the imaginary component of 4z − w.

Im (4z − w) =

w2

4 i–----------=

1

2

4 3 i+( ) 24 i–----------–

4 3 i+( ) 4 i–( ) 2–4 i–

--------------------------------------------

4 13 i+( ) 2–4 i–

-------------------------------

50 4i+4 i–

-----------------

3(50 4i)+

(4 i)–---------------------- (4 i)+

(4 i)+---------------×

196 66i+17

-----------------------

419617

--------- 66i17--------+

56617------

16WORKEDExample

Prove that z1z2 = z–1z–

2.

THINK WRITE

When asked to ‘Prove’ you should not use actual values for the pronumerals. State the general values of z1, z2, z–1 and z–2.

Let z1 = a + biz–1 = a − bi

Let z2 = c + di and z–2 = c − di

Generally, in a proof do not work both sides of the equation at once. Calculate the LHS first.

LHS = (a + bi) × (c + di)LHS = ac + adi + cbi + bdi2

LHS = (ac − bd) + (ad + bc)iLHS = (ac − bd) − (ad + bc)i

Calculate the RHS and show that it equals the LHS.

RHS = (a − bi)(c − di)RHS = ac − adi − bci + bdi2

RHS = (ac − bd) − (ad + bc)iRHS = LHSHence z1z2 = z–1z

–2

1

2

3

17WORKEDExample



Conjugates and divisionof complex numbers

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

2 Graph the following complex numbers and their conjugates on an Argand diagram.a 3 − i b −1 + 3i c −4 − 5i

3 If z = 6 + 3i and w = 3 − 4i, show that = z– .

4 Determine z−1 if z is equal to:

5 If 676z = 10 − 24i, express z−1 in the form x + yi.

6 Express in the form x + yi.

7 Express each of the following in the form x + yi.

8 If z = 2 − i and w = determine each of the following:

9 Write in the form x + yi.

10 Simplify .

11 Determine the conjugate of (5 − 6i)(3 − 8i).

a 7 + 10i b 5 − 9i c 3 + 12id e 2i + 5 f −6 −

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

a b c

d e f

a Re (z + w) b Im (w − z) c Re (z−1 + w−1)d Im (3z + 2w) e Re (4w − 2z)

rememberIf z = a + bi and w = c + di, for a, b, c, d ∈ R, then:

1. The conjugate of z is z– = a − bi.

2. .zw---- ac bd+

c2 d2+------------------ bc ad–( )i

c2 d2+-------------------------+=

remember

2CMathcad

Complex 2

WORKEDExample

127 3i– i 11

WORKEDExample

13

z w– w–

WORKEDExample

143 i 2–

WORKEDExample

15

2 i+3 i–----------

1 i–1 i+---------- 3 2i–

2 3i+-------------- 2 5i+

4 3i–--------------

4 3i–5 2i+-------------- 4 5i–

2 7i–-------------- 2 i 3+

5 i 2–----------------------

WORKEDExample

16

13 i+----------

2 i+1 i+---------- 9 2i–

2 i–-------------- 7 i+

1 i–----------+ +

2 5i+( )2 5i 2–( )3 4 7i+( ) 2 5 8i+( )–---------------------------------------------------



12

If z = 5 − 12i, w = −9 − i and u = 15 − 6i, then:a Re (z−1) is:

b Im is equal to:

c The expression is equal to:

13 If z = 6 + 8i and w = 10 − 3i:a show that = × b generalise the result by letting z = a + bi and w = c + di.

14 Use the result = × to prove that n = ( )n.

15 If z = 4 + i and w = 1 + 3i

a show that

b generalise the result by letting z = a + bi and w = c + di.

16 If z = −5 − 4i and w = 2i, calculate Re (z + w).

17 If z1 = 2 + 3i, z2 = −4 − i and z3 = 5 − i find:

18 If z1 = a + bi and z2 = c + di show that (z1z2)−1 = z1

−1z2−1.

19 a If z = 1 + i find z4, z8 and z12.b Deduce from your results in a that z4n = (2i)2n, n ∈ N.

20 If z = x + yi find the values of x and y such that .

21 Find values for a and b so that z = a + bi satisfies .

22 If z = x + yi, determine the values of x and y such that .

23 If z = 2 − 3i and w = 1 − 2ia find i zz– ii ww–

b Show that iii = +

iii = z– ×

iii =

Find

c i ii d z2 + w2 e z + zw f z–1w–1

A 5 B 12 C D E

A −33 B 103 C 113 D 70 E 0

A 26 − 7i B 64 + 41i C 46 − 29i D 34 − 41i E −64 − 19i

a 2z1 − z2 − 4z3 b z1 2 + z2 3 c − 1 2 3


12169--------- 5

169--------- 12i

169---------

zw( )

2z w– 3u+

zw z w

SkillSH

EET 2.2 zw z w z z

zw----

zw----=

w z

z z z1z2z3 z z z

z 1–z 1+----------- z 2+=

z i+z 2+----------- i=

z 3 4i+=

WORKEDExample

17z w+ z w

zw w

zw----

zw----

1z--- 1

w----



Radians and coterminal anglesWhen a complex number is expressed in a geometrical representation, we use adirected line segment which has length (modulus) and which lies in a certain directionwith respect to the positive Real axis. This angle formed with the positive Real axis iscalled the argument. The argument of a complex number z is written as arg(z) andarg(z) = θ.

Before we look at complex numbers in polar form (in the next section), a new unit ofmeasuring angles is needed, the radian.

Radian measureA radian is the angle subtended by an arc the length of the radius of a circle, as shown in the diagram on the right.

Because the circumference of a circle is given by c = 2πr, there are 2π radians in one complete circle. Taken in an anticlockwise rotation from the positive end of the x-axis as shown, the common angles have radian equivalents.

Therefore, if 2π radians = 360°, 1 radian = g 57.3°.

So an arc of 2πr subtends an angle of 2π radians.

Coterminal anglesConsider the angle 420°. This angle is made up of a full revolution, 360° plus 60°. When using degrees as our unit of angle measure, 420˚ and 60˚ are said to be coterminal angles; that is, angles which differ by a multiple of 360°.

Radians and coterminal angles

1 Draw a circle with a set of axes through its centre. Mark the following on the circum-ference of the circle.

a 0, , π, , 2π b , , , c , ,

2 Convert the following common angle measures to radians.a 45° b 60° c 135° d 270° e 150°

3 Convert the following radian measures to degrees.

a b c d

4 Draw the following sets of coterminal angles:a 30°, 390° b 60°, 420° c 135°, 495°

d , e , f ,

θ

r

r

r

1 radian

360°2π

-----------

20π

π

3 —2π

–4π

–2π

2D

π2--- 3π

2------ π

4--- 3π

4------ 5π

4------ 7π

4------ π

6--- 5π

6------ 7π

6------

7π6

------ 5π4

------ 4π3

------ 5π3

------

π6--- 13π

6--------- 11π

6--------- 23π

6--------- 5π

4------ 13π

4---------



1. To select complex number mode, press andselect Radian mode; scroll down to a + bi and press

.2. To perform simple algebra on complex numbers, use

standard HOME screen operations.

3. To find the real part, imaginary part and conjugate

of a complex number, press , select CPX andthe menu item required — then type in the complex

number, close brackets and press .

4. For a simple complex number, these menu itemsshould not be needed. However, they can be usefulin more complicated questions.

More detailed graphics calculator notes forcomplex numbers are given on pages 114–116.

Complex numbers in polar formThe modulus of zThe magnitude (or modulus or absolute value) of the complex number z = x + yi is the length of the line segment joining the origin to the point z. It is denoted by

z, x + yi or mod z.The modulus of z is calculated using Pythagoras’ theorem.

z = so that we have .

Graphics CalculatorGraphics Calculator tip!tip! Simple algebra of complex numbers

MODE

ENTER

MATH

ENTER

Im (z)

0Re (z)

P(x, y)z = x + yi

y

y

xx

θ

z= x2 + y2

x2 y2+ zz z 2=

Find the modulus of the complex number z = 8 − 6i.THINK WRITE

Calculate the modulus by rule.Because it forms the hypotenuse of a right-angled triangle, the modulus is always greater than or equal to Re (z) or Im (z).

z = = = 10

82 6–( )2+100

18WORKEDExample



The argument of zThe argument of z, arg (z), is the angle measurement anticlockwise of the positive Real axis.

In the figure at right, arg (z) = θ, where

sin θ = and cos θ = or tan θ =

As seen in the previous section, for non-zero z an infinite number of arguments of z exist since, for a given z {θ :θ = ±2nπ, n ∈ N} also represents the position of point P in the figure at right because a clockwise or anticlockwise rotation consisting ofmultiples of 2π radians (or 360°) merely moves P to its original position.

If z = 4 + 2i and w = 7 + 6i, represent the position of w − z on an Argand diagram and calculate w − z.THINK WRITE/DRAW

Calculate w − z. w − z = 7 + 6i − (4 + 2i)= 3 + 4i

Represent it on an Argand diagram as a directed line segment OP.

Use Pythagoras’ Theorem to determine the length of OP.

So w – z = 5

1

2

w

w – z

z

1 2 3 4 5 6 70

123456

Im (z)

P

Re (z)O

3 OP2 32 42+ 25= =OP 5=

19WORKEDExample

Represent z1 = 2 + 3i, z2 = 5 − 2i and z3 = −4 − 2i on the complex number plane and calculate the area of the shape formed when the three points are connected by straight line segments.THINK WRITE/DRAW

Show the connected points on the complex number plane.

Calculate the area of the triangle obtained. The length of the base and height can be found by inspection (base = 9, height = 5).

Area of triangle = × 9 × 5Area of triangle = 22.5 square units.

1

–4–3–2 4 5321

–3–2–1

1234

Im (z)

Re (z)

z1

z2z3

212---

20WORKEDExample

Im (z)

0 Re (z)

P(x, y)z = x + yiy

y

xx

θ

zyz

-------- xz

-------- yx--



To ensure that there is only one value of θ corresponding to z we refer to the principalvalue of θ and denote it by Arg (z). Note the capital A.

Arg (z) is the angle θ in the range −π < θ ≤ π or (−π, π].

Some useful triangles

cos = , sin = , tan = 1

cos = , sin = , tan =

cos = , sin = , tan =

It will be easier if you remember these 2 triangles only — not the ratios shown above. Draw a quick sketch and work out each trigonometric ratio when you need to.

–ππ

1

2√ 3

–6π

–3π

1

1√ 2

–4π

–4π

π4--- 1

2------- π

4--- 1

2------- π

4---

π3--- 1

2--- π

3--- 3

2------- π

3--- 3

π6--- 3

2------- π

6--- 1

2--- π

6--- 1

3-------

Find the Argument of z for each of the following in the interval (−π, π].

a z = 4 + 4i b z = 1 −

THINK WRITE/DRAWa Plot z.

Sketch the triangle that has sides in this 1:1 ratio.

a

From the diagram

θ =

∴ Arg (z) =

This result can be verified using an

inverse trigonometric ratio, θ = tan–1 .

θ = tan–1

θ =

3i

1

2

1

1√ 2

–4π

–4π

4

4

θ

Im (z)

Re (z)

π4---

π4---

3yx--

44---

π4---

21WORKEDExample



THINK WRITE/DRAWb Plot z.

Sketch the triangle that has sides in this ratio.

b

From the diagram

θ =

Arg (z) = −

This result can be verified using an

inverse trigonometric ratio, θ = tan–1 .

θ = tan–1

θ = −

12

1

2√ 3

–6π

–3π

1θ

√ 3

Im (z)

Re (z)

π3---

π3---

3

yx--

3–1

----------

π3---

Convert each of the following into Arguments. a b –

THINK WRITE/DRAWa Sketch the angle. a

Since the given angle is positive, subtract multiples of 2π until it lies in the range (−π, π].

Arg (z) =

= –

b Sketch the angle. b

Since the given angle is negative, add multiples of 2π until it lies in the range (−π, π].

Arg (z) = –

= –

7π4

------ 5π2

------

1 Im (z)

Re (z)–4

π7 —4π

–

27π4

------ 2π–

π4---

1 Im (z)

Re (z)

π

–2

π–

5 —2

–

25π2

------ 2π+

π2---

22WORKEDExample



Find the modulus and Argument for each of the following complex numbers.

a bTHINK WRITE/DRAWa Plot z.

This triangle has sides in the same ratio as a

z = 2

Arg (z) =

These results can be verified by calculating the modulus.

z =

= = 2

θ = tan–1 =

b Plot z. b

Find the modulus. z =

z =

z =

The triangle in the third quadrant will be used to find α but the answer will be finally expressed as θ and Arg (z).

α = tan–1

= tan–1

= 0.955θ = −π + 0.955

Remember Arg (z) can be thought of as the quickest way to get to z.

Arg (z) = −2.187

3 i+ 1– 2i–

1

2

1

2√ 3

–6π

–3π

1θ

√ 3

Im (z)

Re (z)

π6---

3 3( )2

12+

4

1

3------- π

6---

1

1θ

√ 2

Im (z)

αRe (z)

2 1–( )2 2–( )2

+

1 2+

3

3yx--

2–1–

----------

4

Arg (z)

z

23WORKEDExample



Expressing complex numbers in polar formSuppose z = x + yi is represented by the point P(x, y) on thecomplex plane using Cartesian coordinates.

Using the trigonometric properties of a right-angledtriangle, z can also be expressed in polar coordinates asfollows. We have:

cos θ = or x = r cos θ

sin θ = or y = r sin θ

where z = r = and θ = Arg (z).

The point P(x, y) in polar form is shown at right.Now z = x + yi in Cartesian form becomes

z = r cos θ + r sin θ i (after substitution of x = r cos θ, y = r sin θ)

= r (cos θ + i sin θ)

= r cis θ, where cis θ is the abbreviated form of cos θ + i sin θ.

(Note: The acronym ‘cis’ is pronounced ‘sis’.)

Im (z)

0

Re (z)

P(x, y)y

y

x

θ

r = z

xr--

yr--

x2 y2+

Im (z)

0O Re (z)

y

θ

r sinθ

rcosθ

rsin )θP(rcos ,θ

r

Express each of the following in polar (or cis) form.

a z = 1 + i b z = 1 −

Continued over page

THINK WRITE/DRAW

a Plot z. a

The ratio of the sides of this triangle matches the following special triangle:

From the diagramz =

θ =

3i

1

1

1

Im (z)

Re (z)

2

1

1√ 2

–4π

–4π

2π4---

24WORKEDExample



THINK WRITE/DRAW

These results can be verified by rule:

ii Find the value of r using .

r =

=

ii Determine tan θ from tan and hence find θ.

tan θ =

= 1

The angle θ is in the range (−π, π], which is required.

θ =

Substitute the values of r and θ inz = r cos θ + r sin θi = r cis θ.

z =

z =

b Sketch z. b

The ratio of the sides of this triangle is the same as that in the following special triangle:

From the diagramr = 2

θ =

Arg z = −

These results can be verified by rule:

ii Calculate the value of r. r = r = 2

ii Determine the appropriate value of θ. tan θ =

=

=

Substitute for r and θ inz = r cos θ + r sin θi and write in the form r cis θ.

z =

=

3

r z x2 y2+= =12 12+2

θ yx--= 1

1---

4π4---

5 2 cosπ4--- 2 sin

π4---i+

2 cisπ4---

1

1θ

√ 3

Im (z)

Re (z)

2

1

2√ 3

–6π

–3π

π3---

π3---

3

1 ( 3)2+

–3

1-------

3–π3---–

4 2 cos π3---–

2 sin π3---–

i+

2 cis π3---–



103

Express 3 cis in Cartesian form.

THINK WRITE

Sketch z.

Express 3 cis in Cartesian form.

Simplify using exact values from the following triangle:

3 cis =

= 3 × + 3 × i

=

=

π4---

1

3

π–4

Im (z)

Re (z)

2π4---

3

1

1√ 2

–4π

–4π

π4--- 3 cos

π4--- 3 sin

π4---i+

1

2------- 1

2-------

3

2------- 3

2-------

i+

3

2------- 1 i+( )

25WORKEDExample

History of mathematicsABRAHAM DE MOIVRE (26 May 1667 – 27 November 1754)

During his lifetime. . .Christopher Wren

finishes St Paul’s Cathedral.Blackbeard the pirate is killed.The first successful appendectomy is

performed.People are put to

death, as witches, in Salem.

Abraham De Moivre was born in the French town of Vitry but from the age of eighteen he lived in England. The son of a doctor, he was educated at the Protestant Academy at Sedan and then attended college in Paris. In 1685 his family emigrated to England to escape the

growing tensions between Catholics and Protestants in France.

De Moivre contributed to the development of analytic geometry and to the theory of probability. One of his most famous books, The Doctrine of Chances, was published in 1718 and contained major advances in probability theory. In 1725, after investigating mortality statistics, he published Annuities on Lives. Insurance companies of the day used his work to calculate the probabilities of various events. He is best known to students for his formula (r cis θ)n = rn cis nθ which can be used to work out the powers of complex numbers.

It is said that De Moivre was inspired to further research by reading Isaac Newton’s book Principia. He had little spare time so he tore out pages and carried them around with

MQ Maths C Yr 11 - 02 Page 103 Friday, October 12, 2001 4:06 PM


Complex numbers in polar form

In the following questions give arg (z) or Arg (z) correct to 3 decimal places where theangle cannot be easily expressed as a common multiple of π.

1 a Represent z = 4 + 8i on an Argand diagram.b Calculate the exact distance of z from the origin. (Do not use your calculator.)

2 Find the modulus of each of the following.

a z = 5 + 12i b z = c z = −4 + 7i

d z = −3 − 6i e z = f z = (2 + i)2

him, studying them in any free moment. Later in life he became involved in the controversy about whether Newton or Leibniz had been the first to discover calculus. He was appointed by the Royal Society to the commission set up to investigate the rival claims.

De Moivre always had difficulty earning money, but was able to eke out a living by working as a private tutor and by writing books. Unlike many other mathematicians of the time, he could not find a rich patron to support him because he was a foreigner. Even though he was made a fellow of the Royal Society in 1697 and had famous friends such as Newton and Halley, he was always poor and eventually died in poverty.

Apparently, De Moivre predicted the time of his own death. Near the end of his life he

noticed that he needed to sleep for an extra 15 minutes each night. He calculated the date when the cumulative result of this would mean that he was asleep for 24 hours. He died in his sleep on that day.

Questions1. What was the subject of De Moivre’s

book The Doctrine of Chances?2. Why couldn’t De Moivre find a patron?3. How did De Moivre make a living?4. Which famous mathematician played a

major role in his life?5. What was unusual about the date he

died?

ResearchInvestigate how insurance companies use probability to work out how much each insurance policy costs you.

remember1. The magnitude (or modulus or absolute value) of z = x + yi is the length of the

line segment from (0, 0) to z and is denoted by z, x + yi or mod z.

2. z = and z = .

3. arg z = θ where tanθ = . ∴ θ = tan–1

4. z × in, n ∈ N produces an anticlockwise rotation of 90n degrees.5. z = r cos θ + r sin θi = r cis θ in polar form.6. Arg z is the angle θ in the range −π < θ ≤ π.

x2 y2+ z z 2

yx-- y

x--

remember

2E

Mathca

d

Complex 1

WORKEDExample

18 5 2i–

3 2i+


C h a p t e r 2 N u m b e r s y s t e m s : c o m p l e x n u m b e r s 1053 If z = 3 + i, w = 4 − 3i and u = −2 + 5i then:

i represent each of the following on an Argand diagramii calculate the magnitude in each case.

4 a Show the points z1 = −3 + 0i, z2 = 2 + 5i, z3 = 7 + 5i and z4 = 9 + 0i on the complexnumber plane.

b Calculate the area of the shape formed when the four points are connected bystraight line segments in the order z1 to z2 to z3 to z4.

5 a Show the points z = −1 + 3i, u = 3 and w = 3 + 12i on the complex number plane.b Calculate the area of the triangle produced by joining the three points with straight

line segments.

6 Find the argument of z for each of the following in the interval [0, 2π]. (Give exactanswers where possible.)

7 Convert each of the following into Arguments.

8 Find the modulus and Argument of each of the following complex numbers.

9 Express each of the following in the polar form z = r cis θ where θ = Arg (z).

10 Express each of the following complex numbers in Cartesian form.

11If z = 3 − 50i and w = 5 + 65i the value of z + w is:

a z − w b u + z c w − ud w + z e z + w − u f z2

a z = 3 + 2i b z = c z = 5 − 5i d z = −4 + 8i

e z = −2 − f z = 6 − 10i g z = 3i h z = i z = −6i j z = 55

a b − c d −

e f g − h −

a 3 − 3i b −5 + 5i c −1 − de −7 − 10i f 6i − 2 g

a z = −1 + i b z = c z = d z =

e z = − f z = −

a 2 cis b 3 cis c d 4 cis

e f 8 cis g cis π

A 64 B 15 C 17 D 225 E 289

WORKEDExample

19

WORKEDExample

20

WORKEDExample

213 i+

2 3i 7–

WORKEDExample

22 3π2

------ 11π6

--------- 15π8

--------- 5π4

------

19π6

--------- 20π7

--------- 18π5

--------- 13π12

---------

WORKEDExample

23 3i 4 3 4i+

3 i+( )2

WORKEDExample

24 6 2i+ 5– 5i– 5 15i–

12---

32

-------i–14--- 1

4---i+

WORKEDExample

25 2π3

------ π4--- 5 cis

5π6

------ −π3---

7 cis −7π4

------ π

2--- 3




12The perimeter of the triangle formed by the line segments connecting the points2 − 4i, 14 − 4i and 2 + i is:

13

The Argument of is:

14In polar form, 5i is:

15

The Cartesian form of cis is:

Basic operations on complex numbers in polar form

Addition and subtractionIn general there is no simple way to add or subtract complex numbers given in the polarform r cis θ. For addition or subtraction, the complex numbers need to be expressed inCartesian form first.

A 13 B 30 C 10 D 17 E 25

A B C D E

A cis B cis 5π C cis D 5 cis 5π E 5 cis

A B C D E

Multiplication in polar formIn earlier sections we performed multiplication and division on complex numbers in standard form. This is quite a lengthy process for both these operations. However, as is the case in many aspects of mathematics, patterns exist that make the job so much easier. Work through the following investigation that will form the basis of future work.

1 Given that z = 1 + i and w = 2 + 2i: a find zw in standard formb express the product from part a in polar formc verify that | zw | = | z || w |d verify that arg (zw) = arg (z) + arg (w).



4 3 4i–

π6--- π

3--- 5π

6------ π

6---– π

3---–


π2--- 5π

2------ π

2---


WorkS

HEET 2.1 3 –7π6

------

12--- 3

2-------i+ −1

2--- 3

2-------i+ − 3

2------- 1

2---i+ −3

2--- 3

2-------i+ − 3

2------- 1

2---i–

3



2 Given that z = 1 – i and w = 2 – 2 i:a find zw in standard formb express the product from part a in polar formc verify that | zw | = | z || w |d verify that arg (zw) = arg (z) + arg (w)

3 Given that z = a + bi and w = c + dia find zw in standard formb verify that | zw | = | z || w |c For z = r1 cis θ = r1 (cos θ + i sin θ) and w = r2 cis φ = r2 (cos φ + i sin φ),

show that zw = r1r2[(cos θ cos φ – sin θ sin φ) + i(cos θ sin φ + sin θ cos φ)]

4 Using the trigonometric identities:cos (A + B) = cos A cos B – sin A sin Bsin (A + B) = cos A sin B + sin A cos Bverify that zw = r1r2[cos(θ + φ) + i sin (θ + φ)]

= r1r2 cis (θ + φ)

This investigation illustrates the following useful facts concerning multiplication of complex numbers in polar form:

If z and w are two complex numbers, then| zw | = | z || w | and arg (zw) = arg (z) + arg (w)

Similarly, for division of complex numbers:

= and arg = arg (z) – arg (w)

The proofs required to establish these rules are outside the Mathematics C syllabus and will not be included in this chapter on complex numbers.

3

zw---- z

w-------- z

w----

Express 5 cis in the form r cis θ where θ ∈ (−π, π].

THINK WRITE/DRAW

Simplify using the multiplication rule zw = r1r2 cis (θ + φ) (see part 4 above).

=

= 10 cis

Sketch this number.

Subtract 2π from θ to express the answer in the required form.

10 cis = 10 cis

π4--- 2 cis

5π6

------¥

1 5 cisπ4--- 2 cis

5π6

------× 5 2×( ) cisπ4--- 5π

6------+

13π12

---------

2 Im (z)

Re (z)

11 —12

π–

13 —12

π

313π12

--------- –

11π12

---------

26WORKEDExample



Express z1z2 in Cartesian form if and .

THINK WRITE

Use z1z2 = r1r2 cis (θ1 + θ2).

Write the result in standard form.

z1z2 =

=

=

= 2 cos + 2 sin i

= 2 × 0 + 2 × 1i

=

z1 2 cis5π6

------= z2 6 cis π3---–

=

1

2

2 cis5π6

------ 6 cis π3---–

×

( 2 6)× cis 5π6

------ π3---–

2 3 cisπ2---

3π2--- 3

π2---

3 3

2 3i

27WORKEDExample

If and express the product zw in polar form.

THINK WRITE/DRAW

Sketch z.

Write z in polar form. Use the special triangle below:

The ratio of sides in z is 5 times that of the sides in this triangle.

Let z = r1cis θ1.r1 = 5 × 2 = 10

θ1 =

z 5 3 5i+= w 3 3 3i+=

1

5θ

√ 35

Im (z)

Re (z)

2

1

2√ 3

–6π

–3π

π6---

28WORKEDExample



THINK WRITE/DRAW

Verify this by rule if you wish.

θ = tan–1 , so

Therefore z =

Sketch w.The ratio of sides in w is 3 times that of the sides in the triangle shown in step 2.

Verify this by rule if you wish. r2 = 3 × 2 = 6

θ2 =

Write w in polar form. Let w = r2 cis θ2

Then

tan , so

Therefore

Determine zw usingz1z2 = r1r2 cis (θ1 + θ2).

zw =

=

=

3r1 (5 3)2 52+ 10= =

5

5 3---------- θ1

π6---=

10 cisπ6---

4

3θ

3√ 3

Im (z)

Re (z)

5π3---

6

r2 32 (3 3)2+ 6= =

θ23 3

3---------- 3= = θ2

π3---=

w 6 cisπ3---=

7 10 cisπ6--- 6 cis

π3---×

60 cisπ6--- π

3---+

60 cisπ2---

Express 10 cis cis in the form r cis θ where θ ∈ (−π, π].

Continued over page

THINK WRITE/DRAW

Simplify using the division rule. =

=

–π3---

5÷ 5π6

------

1 10 cis π3---–

5 cis5π6

------÷ 2 cis π3---– 5π

6------–

2 cis 7π6

------–

29WORKEDExample


110


Powers of complex numbers

Whole powers of

z

As with real numbers, powers of complex numbers can be written as:

z

n

=

z

×

z

×

z

×

z

× …

×

z

to

n

factors.Since

z

=

a

+

bi

is a binomial (containing two terms) we can express

z

n

using Pascal’sTriangle to generate the coefficients of each term.

11 1

1 2 1 1 3 3 1 1 4 6 4 1

5th row

→

1 5 10 10 5 1 and so on.(

a

+

bi

)

5

can therefore be expanded using the elements of the fifth row of Pascal’sTriangle:

(

a

+

bi

)

5

= 1

a

5

+

5

a

4

(

bi

)

1

+

10

a

3

(

bi

)

2

+

10

a

2

(

bi

)

3

+

5

a

(

bi

)

4

+

(

bi

)

5

(

a

+

bi

)

5

= 1

a

5

+

5

a

4

bi

+

10

a

3

b

2

i

2

+

10

a

2

b

3

i

3

+

5

ab

4

i

4

+

b

5

i

5

(

a

+

bi

)

5

= 1

a

5

+

5

a

4

bi

−

10

a

3

b

2

−

10

a

2

b

3

i

+

5

ab

4

+

b

5

i

(

a

+

bi

)

5

= 1

a

5

−

10

a

3

b

2

+ 5

ab

4

+

5

a

4

bi

−

10

a

2

b

3

i + b5i(a + bi)5 = 1a5 − 10a3b2 + 5ab4 + (5a4b − 10a2b3 + b5)i grouped into standard form.

Re [(a + bi)5] = 1a5 − 10a3b2 + 5ab4

Im [(a + bi)5] = 5a4b − 10a2b3 + b5

The coefficients of each term of the expansion of (a + bi)n can be found using yourgraphics calculator. For example, the coefficients of the expansion of (a + bi)5 can alsobe written as:

5C0a5 + 5C1a4(bi)1 + 5C2a3(bi)2 . . . 5C5(bi)n

where 5C0, 5C1, . . .,

5C5 represent the coefficients.To find 5Co: type , press , select PRB, 3:nCr, press

to obtain 1. Any coefficients can be found using this method.

THINK WRITE/DRAW

Sketch this number.

State θ, the principal argument. Arg (z) =

State the result in polar form. Arg ()z = 2 cis

2

7 —6

5 —6

2Im (z)

Re (z)

π–

π–6π

35π6

------

45π6

------

Graphics CalculatorGraphics Calculator tip!tip! Pascal’s Triangle coefficients

5 MATH ENTER 0 ENTER



Negative powers of z

Your earlier studies have shown that z–1 = . Similarly, z–3 = .

Fractional powers of zFractional powers of complex numbers generally follow the same rules as with realnumbers.

= Our discussion here will deal only with the square root of z, where .

Use Pascal’s Triangle to expand (2 - 3i)3.

THINK WRITE

Use the third row of Pascal’s Triangle to expand (1 3 3 1).Use brackets to keep the negative sign of the second term.

(2 − 3i)3

= 1(23) + 3(2)2(−3i) + 3(2)(−3i)2 + (−3i)3

Simplify the expression. = 8 − 36i + 54i2 − 27i3

= 8 − 36i − 54 + 27i= −46 − 9i

1

2

30WORKEDExample

1z--- 1

z3

----

Evaluate (3 - i)-4.

THINK WRITE

First find the expansion with a positive power. Use the fourth row of Pascal’s Triangle to expand (1 4 6 4 1).

(3 − i)4

= 34 + 4(3)3(−i) + 6(3)2(−i)2 + 4(3)(−i)3 + (−i)4

Use brackets to keep the negative sign with the second term.

= 81 − 108i − 54 + 12i + 1

Simplify to obtain the standard form. = 28 − 96i

Express this as the denominator then multiply by the complex conjugate.

(3 − i)–4 = ×

(3 − i)–4 =

(3 − i)–4 =

Write the final expression in standard form.

(3 − i)–4 = + i

1

2

3

41

28 96i–( )------------------------ 28 96i+( )

28 96i+( )-------------------------

28 96i+784 9216+---------------------------

28 96i+10 000

--------------------

57

2500------------ 6

625---------

31WORKEDExample

zpq--- z

pq

z z12---

=



Express in standard form.

THINK WRITE

Let be a complex number suchas (a + bi), where a, b ∈ R.

Let = a + bi

All dialogue given in the ‘write’ column should appear as ‘communication’ in your working.

Square both sides:3 + 4i = (a + bi)2

3 + 4i = a2 + 2abi − b2

Equating real and imaginary terms:3 = a2 − b2 (1)4 = 2ab (2)

a = = (3) from (2)

Substitute for a into (1)

3 = − b2

3 = − b2

3b2 = 4 − b4

0 = b4 + 3b2 − 4= (b2 − 1)(b2 + 4)

Therefore, b2 = 1, b = ±1or b2 = −4, b = ±2iSince a and b are real numbers discard b = ±2i.Substitute for b = ±1 into (3)

a = ± ∴ a = 2 or −2

State the final result in standard form. Therefore = 2 + i or −2 − i= ±(2 + i)

Verify this result. [±(2 + i)]2 = 4 + 4i − 1 = 3 + 4i

3 4i+

1 3 4i+ 3 4i+

2

42b------ 2

b---

2b---

2

4

b2

-----

21---

3 3 4i+

4

32WORKEDExample

remember1. If z1 = r1 cis θ1 and z2 = r2 cis θ2, then:

z1 × z2 = r1r2 cis (θ1 + θ2)

cis (θ1 − θ2).

2. A complex number zn = (a + bi)n can be expanded using Pascal’s Triangle to generate the coefficients of each term.

3. Negative powers of z: z–n =

4. Fractional powers of complex numbers:

z1

z2----

r1

r2----=

1

zn

----

zpq--- z

pq=

remember



113

Basic operations on complex numbers in polar form

1

Express each of the following in the form

r

cis

θ

where

θ

∈

(

−

π

,

π

].

a

2 cis cis

b

5 cis cis

c

6 cis cis

π

d

cis cis

e

cis cis

2

Express the resultant complex numbers in question

1

in standard form.

3

Express the following products in polar form.

4

Express each of the following in the form

r

cis

θ

where

θ

∈

(

−

π

,

π

].

5

If

z

=

and

w

=

2 cis then express each of the following in:

i

polar form

ii

standard form.

6

If

z

=

1

−

i

and

w

=

, write the following in standard form.

7

Determine in standard form.

8

Write in the form

x

+

yi

.

9

a

is equal to:

b

If then is:

c

If and

w

=

2

+

2

i

then is equal to:

a

(2

+

2

i

)(

+

i

)

b

(

−

3

i

)(2

−

2

i

)

c

(

−

4

+

i

)(

−

1

−

i

)

a

12 cis

b

36 cis

c d

e

a

z

3

b

w

4

c

z

4

d

w

5

a

z

−

4

b

w

−

3

c

z

−

3

d

w

−

5

e f

z

2

w

3

A B C D

−

6

i

E

A

1

+

i

B C

(1

−

i

)

D E

A

−

4

+

4

i

B C

2

D

−

4

i

E

−

8

2FGC program

Power ofa complex

number

WORKEDExample

26 π4--- 3× π

2--- 2π

3------ 4× π

3---–

3π4

------ 5×

3 5π6

------– 2× π

2---–

7 7π12------–

2× 5π12------

WORKEDExample

27

WORKEDExample

28 3 3 3 4 3

WORKEDExample

29 5π6

------ 4 cis π3---÷ 3π

4------ 9 cis π

6---–

÷

20 cis π2---–

5 cis π5---–

÷ 4 3 cis 4π7

------ 6 cis 11π14

---------÷

3 5 cis 7π12------–

2 10 cis 5π6

------÷

WORKEDExample

30

3 cis3π4

------ π4---–

WORKEDExample

31

– 3 i+z3

w4------

2 2i+( )2 1 3i–( )4

3 i–( )6

2 2 3i–( )3----------------------------


5 cis π3---–

8 cis π6---–

×

6 2i 2– 10i 6– 3 6 6

z ( 6 2)+ ( 6 2– )i+= 2z 3–

2i164------ 2 i+ –1 2i–

z –1 3i–=w4

z3------

2 3



10 If and , find the modulus and the argument of .

11 If z = 4 + i and w = −3 − 2i, determine (z + w)9.

12 Find z6 + w4, if and w = 2 − 2i.

13 If , and , find the modulus and

the argument of .

14 Express each of the following in standard form:

a b c d

Graphics calculator notes for complex numbers

The following notes are designed to support TI–83 graphics calculators. Two menus areused: (screen shown below left) and (screen at right).

The highlighted terms shown in the screens above should be selected when using and facilities. Press to exit the menu.

To enter complex numbers in rectangular or Cartesian form (using the menu):

Task KeystrokesEnter 4 + 2i then store in A.

Enter 3 − 3i then store in B.

z 2 cis 3π4

------= w 3 cis π6---=

z6

w4------

z 2 2i–=

z1 5 cis 2π5

------– = z2 2 cis

3π8

------= z3 10 cis π12------=

z12 z2

3×

z34

-------------------

WorkS

HEET 2.2WORKEDExample

32 5 12i+ 5 12i– 2 2i+ 3 4i–

MODE MATH

MODE MATH 2nd QUIT

MODE

4 + 2 2nd . STO

ALPHA A ENTER

3 - 3 2nd . STO

ALPHA B ENTER


C h a p t e r 2 N u m b e r s y s t e m s : c o m p l e x n u m b e r s 115To recall complex numbers stored in A and B and perform operations withthem using the menu:

The tabe below shows how to use the facilities:

Task KeystrokesAdd A + B to get 7 − i.

Repeat for subtraction, multiplication, division, powers, using the recall

function .To repeat these operations without using

the function, brackets will be needed around each number. For A + B you would press the keys shown.

To enter 3(4 + 2i)4 without the function press the keys shown.

Repeat a selection of operations from your text.

Task Keystrokes1. Find the conjugate of A and perform

some operations with it and A. Repeat for addition, subtraction, etc.

select

CPX 1:

2. Find the real part of

((4 + 2i) × (4 − 2i)) using the keystrokes shown.Note the double brackets around the whole product.

select CPX 2:

select CPX 1:

3. Find the imaginary part of

((4 + 2i) × (4 − 2i) using the keystrokes shown.

select CPX 3:

select CPX 1:

4. Find the angle formed with the x-axis

(in polar form) given in radians using the keystrokes given. Note that the initial setting is in radians.

select CPX 4:

5. Finds the absolute value or modulus

of the number using .

select CPX 5:

MODE

2nd STO ALPHA A ENTER

+ 2nd STO ALPHA B

ENTER ENTER

RCL

RCL( 4 + 2 2nd .

) + ( 3 - 3

2nd . ) ENTER

RCL 3 × ( 4 + 2

2nd . ) ^ 4 ENTER

MATH

( 2nd STO ALPHA A

ENTER ) × MATH

ENTER 2nd STO ALPHA

A ENTER ) ENTER

MATH ENTER (


) × ENTER


) ) ENTER

MATH ENTER (


) × MATH

ENTER 2nd STO ALPHA A

ENTER ) ) ENTER

MATH ENTER 2nd

STO ALPHA A ENTER )

ENTER

42 22+

MATH ENTER 2nd

STO ALPHA A ENTER )

ENTER



Repeat a selection of exercises using complex numbers in polar form.

Graphing complex numbers on an Argand diagram

The TI-83 graphics calculator has no built-in function to graph on the complex planebut it can be manipulated to produce a similar effect using the facilities.

Note that due to the method used in plotting complex numbers, they must be entered as2 lists: real and imaginary (in rectangular or standard form).

Powers of complex numbers

These can be entered in either standard form or polar form.Example: Evaluate and give the answer in both polar and standard forms.

Rework a selection of similar exercises throughout this chapter using the TI–83graphics calculator.

6. Converts a number given in polar form to rectangular form (using the angle obtained in part 4 and modulus from part 5).

select CPX 6:

7. Converts a number given in rectangular form to polar form.

select CPX 7:

Task KeystrokesSelect plot type. select 1: select On

(second graph) Enter 4 + 2i, the origin, and the conjugate 4 − 2i in two lists. The points are listed in the order they should be connected. Add the real part to list 1 (L1) and imaginary part to list 2 (L2).

To select the lists to be graphed. select OPS 8:

Adjust the scales of the graph. Trace the points on the graph. or to give

coordinates of the points.

Task KeystrokesSet mode to complex rectangular. select a+bi Enter the power of the number as a decimal power.

Result displayed in rectangular form. = 2.197 + 0.910 i

Change display to polar. select CPX 7: = 2.378e^ (0.393i)

Task Keystrokes

4.472 2nd LN 0.464 2nd

. ) MATH

ENTER ENTER


MATH ENTER ENTER

STAT

2nd Y= ENTER

ENTER 2nd QUIT

STAT 1 L1 L2

404

20

-2

2nd STAT ENTER

2nd 1 , 2nd 2 )GRAPH

WINDOW -5 5 1

TRACE �

�

4 4i+

MODE

2nd QUIT ( 4 + 4

2nd . ) ^ .5 ENTER

MATH ENTER ENTER



117

Complex numbers: applications

1

Choose a complex number that falls in the first quadrant of the complex plane. Calculate the first 8 powers of this number and investigate any pattern that exists between the modulus of each of the powers. Plot each power on an Argand diagram. What do you notice?

2

Let

z

=

r

cis

θ

, a complex number. Find, in terms of

r

and

θ

:

a iii

z

2

=

z

.

z

ii

z

3

=

z.z

2

iii

z

4

=

z.z

3

iv

z

5

=

z.z

4

v

z

6

=

z.z

5

vi

z

7

=

z.z

6

b

Write the moduli of the powers of

z

as a sequence.

c

What do you notice about the sequence given in part

b

?

3

As mentioned at the beginning of this chapter the equation

z

2

=

1 has two solutions,

z

= ±

1, whereas the equation

z

=

1 has only one solution,

z

=

1. The

equation

z

3

=

1 has 3 solutions,

z

=

1, cis and cis .

Graph these solutions on an Argand diagram. Express all solutions in both rectangular and mod–arg form.

4

Let

z

=

x

+

yi

. Therefore

|

z

|

=

, and

|

z

|

2

=

x

2

+

y

2

, where this is the general equation of a circle, of radius

|

z

|

, about the origin. Graph this circle and fully label the path of the rotating

z

as it moves about the origin. Therefore, what is the meaning of the statement

|

z

|

<

x

+

yi

. Sketch

|

z

|

< 4 and

|

z

|

> 1.

5

Research the life of William Rowan Hamilton and his contribution to the study of complex numbers.

6

Research the area of mathematics called

fractals

. You will investigate this fascinating area in more detail later in your studies.

7

In chapter 1, you were introduced to the term ‘transcendental numbers’ — irrational numbers that are not algebraic; that is, cannot be produced by the algebraic operations of addition, subtraction, multiplication and division, and by taking roots. Pi (

π

) is one such transcendental number and

e

is another,

where

e

=

e

1

=

1

0

+

+

+

+

. . .

(and 3! = 3

×

2

×

1, and so on. The symbol 3! is referred to as factorial 3.)The function

e

x

is referred to as

the exponential function

.

e

x

=

x

0

+

+

+

+

. . .

The graph of the function

e

x

is especially interesting because the slope of the curve at any point equals the value of the curve, at that point. That is, the slope of a tangent to the curve at

x

= e2 is e2.Euler discovered a special relationship between e and i,

where ei = i0 + + + + . . .

2π3

------ −2π3

------

x2

y2+

11

1!----- 12

2!----- 13

3!----- 14

4!-----

x1

1!----- x

2

2!----- x

3

3!----- x

4

4!-----

i1

1!----- i

2

2!----- i

3

3!----- i

4

4!-----


118


Write 4 expressions for

e

i

, with increasing numbers of terms and simplify them where possible. The alternating positive and negative signs suggest that the expression is approaching a particular value as the number of terms in the series increases. You might find it more methodical to list the results as each new term is added as you ‘creep’ closer to the value. Can you suggest what that value might be? What is the modulus of this number? Use the TI-83 graphics calculator to evaluate

e

i

.

History of mathematicsW I L L I A M R OWA N H A M I LT O N ( 1 8 0 5 – 1 8 6 5 )

During his life . . .Charles Darwin

developed his theory of evolution.Charles Babbage developed the first automatic

digital computer.Gregor Mendel

laid the mathematical

foundation for the science of genetics.

Sometimes considered the second greatest mathematician of the English-speaking world, after Sir Isaac Newton. William Hamilton was born in Dublin, Ireland on August 3, 1805. Even the fact that Hamilton did not attend school before he entered college, did not deter his thirst for knowledge. By the age of three he was skilled at reading and arithmetic, by the age of five he read and translated Latin, Hebrew and Greek, and by the age of 14 he could speak 14 languages.

By the age of 21 he published a paper entitled ‘A Theory of Systems of Rays’, introducing and developing concepts that created the field of mathematical optics.

Propelled by the success of this work, at 22 he was unanimously voted into the chair of the professor of astronomy at Trinity College (Dublin), even though he was still an undergraduate and had not applied for the position.

In 1833 Hamilton further developed his work on complex numbers and in 1843 he released what he considered to be his greatest discovery — the algebra of quaternions. With these ordered sets of four numbers, magnitude and direction in 3-dimensional space could be determined. The fact that multiplication of quaternions is not commutative led to the development of the first ‘ring’ in which the commutative property does not hold. This inspiration came to him while he was crossing the Brougham Bridge in Dublin and he left the inscription: i2 = j2 = k2 = ijk = −1 in a stone in the bridge. A stamp featuring these quaternions was issued in Ireland in 1983.

His work also led to the development of the concepts of vectors, scalars and tensors, which you will encounter later in your studies. Plagued throughout his life with alcoholism, he died in 1865.

Research1. Find out more about quaternions.2. Research the notion of ‘rings’.

MQ Maths C Yr 11 - 02 Page 118 Friday, October 12, 2001 4:07 PM


Introduction to complex numbers• We define the ‘imaginary number’ i as having the property that i = .• A complex number z = x + yi with x, y ∈ R and C = {z: z = x + yi, x, y ∈ R} defines

the set of complex numbers.• The real part of z is x and is written as Re (z).• The imaginary part of z is y and is written as Im (z).

Basic operations using complex numbers• If z and w are two complex numbers such that z = a + bi and w = c + di for a, b, c,

d ∈ R then:1. z = w if and only if (i.e. iff) a = c and b = d2. z + w = (a + c) + (b + d)i3. z − w = (a − c) + (b − d)i4. kz = ka + kbi, for k ∈ R5. z × w = (ac − bd) + (ad + bc)i.

Conjugates and division of complex numbers• If z = a + bi:

1. The conjugate of z is = a − bi.2. z. = a2 + b2.

The polar form of complex numbers• The magnitude (modulus or absolute value) of z = x + yi is the length of the line

segment from (0, 0) to z. It is denoted by z, x + yi or mod z.• z = and z = .• The argument of z, arg (z), is the angle measurement anticlockwise of the positive

Real axis and arg (z) = θ where θ = tan–1 .

• z = x + yi can be expressed in polar form as z = r cos θ + r sin θi = r cis θ.• Arg (z) is the angle θ in the range −π < θ ≤ π and is called the principal argument.

Basic operations on complex numbers in polar form• If z1 = r1 cis θ1 and z2 = r2 cis θ2, then:

1. z1 × z2 = r1r2 cis(θ1 + θ2)

2. cis(θ1 − θ2)

summary1–

z zz

x2 y2+ z z 2

yx--

z1

z2----

r1

r2----=



Questions 1 and 2 refer to the complex number .

1

The real and imaginary parts of z respectively are:

2

The Argand diagram which correctly represents z is:

3 Simplify i6 − i3 (i2 − 1).

Questions 4 and 5 refer to the complex numbers u = 5 − i and v = 4 + 3i.

4

The expression 2u − v is equal to:

5

The expression uv is equal to:

6

If z = 5 − 12i, decide which statement is true concerning −iz.A −iz = B −iz = 12 − 5iC The point z is rotated 90° clockwise.D Re (−iz) = 0E Im (−iz) = −i

A and 4 B and −4 C 4 and D −4 and E and −4i

A B C

D E

A 1 − 4i B −3 − 7i C 6 − 5i D 5 + 8i E 14 + i

A 9 + 2i B 20 − 3i C 20 + 3i D 15 − 4i E 23 + 11i

CHAPTERreview

z 2 5 4i–=


2A 2 5 2 5 2 5 2 5 2 5


2AIm (z)

0

4

Re (z)52

zIm (z)

0

–4

Re (z)52

z

Re (z)

0

–4

Im (z)52

z

Re (z)

0

4

Im (z)52

zIm (z)

0–4 Re (z)

52z

2A

2B mmultiple choiceultiple choice


2B


2B 13


C h a p t e r 2 N u m b e r s y s t e m s : c o m p l e x n u m b e r s 1217 If z = 3 − 8i, then find:

8 If z = 2 − 5i, u = −3 + i and w = 1 + 2i evaluate:

9

Im is equal to:

10

If z = 3i and w = 4 − i then z is equal to:

11

The expresion simplifies to:

12 If z = 6 − 2i and w = 5 + 3i, express in the form a + bi, a, b ∈ R.

13

Arg (2 − 2i) is equal to:

14

The polar form of −3 + 3i is:

15 If z = −7 − 7i, express z in polar form.

16

How many degrees apart are two consecutive roots of z8 = 1 on the unit circle?

17

If z1 = 10 cis and z2 = 5 cis then z1z2 in polar form is:

a Im (z2) b a and b if z3 = a + bi.

a z − 2u + 3w b z c uz + w

A 2 B − C D E −2

A 12 + 3i B 12 − i C 7 + 3i D 12 − 3i E z = 4 − 2i

A B 3 + 7i C D 4 − 2i E 3 − i

A π B C D E 2π

A B C D E

A 180 B 90 C 135 D 225 E 45

A 50 cis B 15 cis C 2 cis D 15 cis E 2 cis

2B

2B,C


2C1 2i+1 i–

--------------

12--- 3

2--- 2

3---


2Cw

2Cmmultiple choiceultiple choice

2i1 i+----------- 3

2 i–----------–

15---–

25---i+ 1

4---

34---i–

2Czw----


2Eπ4--- 3π

4------ π

4---–

2Emmultiple choiceultiple choice

3 2 cisπ4--- 3 2 cis

3π4

------ 3 cis3π4

------ 3 cis π4---–

3 2 cis5π4

------

2Emmultiple choiceultiple choice

2E


2Fπ4--- π

6---–

π12------ 5π

12------ π

12------ π

12------–

5π12------–



18

In standard form, is equal to:

19

in standard form is:

20

The solutions to z3 = in polar form are:

A 4 + 4i B −4 − 4i C 4 − 4i D −4 + 4i E 36 − 36i

A B C

D E

A 2 cis , 2 cis , 2 cis B , ,

C , , D 4 cis

E ,


2F 12 2 cis3π4

------ 3 cis π2---–

÷


2F 5 5 3i–

5 5 3i –( )± 5 3 5i– 5 5 3 i+( )±

302

----------102

---------- i– ± 15 5i –

2F mmultiple choiceultiple choice

testtest

CHAPTERyyourselfourself

testyyourselfourself

2

3 i+π6--- 5π

6------ 3π

2------ 2 cis

π6--- 2 cis

5π6

------ 2 cis π2---–

23 cisπ18------ 23 cis

13π18

--------- 23 cis 11π18

---------– π

6---

2 cisπ3--- 2 cis 2π

3------–


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