nj complex numbers lecture notes student edition

2011 Senior High 2 H2 Mathematics NJC Mathematics Department

Complex Numbers: Algebra and Loci OBJECTIVES

At the end of this topic, students should be able to

understand that i = −1 is a root of 2 1 0x + = , and that a quadratic equation has two real roots or two complex roots.

recognise that the set of real numbers is a subset of the complex numbers carry out addition, subtraction, multiplication and division of complex numbers given

in Cartesian form [x + iy, x, y ∈ ], including finding the square root of complex numbers.

find the modulus and argument of a complex number given in Cartesian form. use the relation zz* = |z|2 to calculate the modulus and the reciprocal of a complex

number. solve simple equations involving a complex variable by equating real parts and

imaginary parts. understand that roots of polynomial equations with real coefficients are either real, or

occur in complex conjugate pairs. solve polynomial equations with real coefficients when a root (either real or complex)

is given. represent complex numbers expressed in Cartesian form by points in the Argand

diagram, and interpret the terms ‘real part’, ‘imaginary part’, ‘modulus’, ‘argument’ and ‘conjugate’ geometrically.

convert a complex number from Cartesian form [x + iy, x, y ∈ ] to polar form [r(cos θ + i sin θ), r > 0, −π < θ ≤ π], exponential form [ reiθ ] and vice versa. carry out multiplication and division of two complex numbers given in polar form. represent complex numbers by points in the Argand diagram, and interpret

geometrically the effects of adding/subtracting and multiplying/dividing two complex numbers.

interpret the multiplication by i as ‘rotation by 90˚ anti-clockwise’ in the Argand diagram.

use De Movire’s Theorem with rational indices to find the powers and nth roots of a complex number.

sketch the loci of simple equations and inequalities in the Argand diagram.

Important Note:

Throughout this set of notes, the symbol indicates that the graphic calculator (GC) can be used to solve the problem.

Parts where you have to attempt are indicated with symbol.

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§0 A Brief Complex History

History of complex numbers began sometime in the 16th century when Italian mathematician Gerolamo Cardano was considering cubic and quartic polynomials and realised that sometimes to solve them, the manipulation of square root of negative numbers was required. About 30 years later, this idea was further developed by hydraulic engineer Rafael Bombelli, who assumed that numbers of the form a b+ − existed. This laid further groundwork for development of complex numbers.

From then, there were more works which made use of these complex numbers and through these works, much more was discovered. However, it was only during the time of Leonard Euler when he started the use of the alphabet i to represent 1− .

What followed was the Argand diagram in 1806 named after Jean Robert Argand where complex numbers were represented geometrically.

Complex numbers continued to develop after this. Work on them led to the fundamental theorem of algebra and a branch of mathematics called complex analysis. They are also widely used in the study of physics such as in quantum mechanics and electric circuitry.

§1 Definition and Terminology

A complex number, z ∈ , is a number of the form z = x + iy , where x, y ∈ .

The imaginary unit, i, takes the value of 1i −= .

x is the real part of z and is denoted by Re(z). (i.e. ( ) xz =Re ) y is the imaginary part of z is denoted by Im(z). (i.e. ( ) yz =Im ) [The symbol is used to denote the set of complex numbers.]

In particular, if , then z = iy is a purely imaginary number. 0=x

if , then z = x is a real number. 0=y Hierarchy of the number system:

Complex Numbers

Real Numbers Purely Imaginary Numbers

Rational Numbers Irrational Numbers

Integers Fractions

Negative Integers Zero Positive Integers

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§2 Equality of Complex Numbers

Given that 111 iyxz += and 222 iyxz += where 1 2 1 2, , ,x x y y ∈R , then

21 zz = ⇔ 21 xx = and 21 yy = .

In particular, 0and00 ==⇔= yxz .

Example 1 Find x and y, where , if ,x y∈R i24)3(i2 +=−++ yxyx .

Solution Comparing real part on both sides, we get 2 4 (1x y )+ = − − −

Comparing imaginary part on both sides, we get 3 2x y (2)− = − − −

Solving equations (1) and (2) simultaneously, we get 87

x = and 107

y = .

Note: We must ensure that the terms on both sides of the simultaneous equations are

real. §3 Algebraic Operations on Complex Numbers 3.1 ‘i’ follows all operations in algebra.

i 1= −

(i) Addition: 3i + 4i = 7i (ii) Subtraction: 10i – 3i = 7i (iii) Multiplication: ; iii aa =× baba ii)( = ++ ; 2i i ia b ab ab× = = − , where a, b ∈ .

2i = −1 i i= = − 1 ; i i ; ( )3 2 4 2 2i i i= ⋅ = ; etc…

(iv) Division: 12

1 1 i ii ii i i i

− = = ⋅ = = −

For sections 3.2 to 3.4, consider two complex numbers 111 iyxz += and 222 iyxz += where . 1 2 1 2, , ,x x y y ∈R

3.2 Addition & Subtraction of Complex Numbers

(a) ( ) ( )1 2 1 1 2 2 1 2 1 2i i ( ) i ( )z z x y x y x x y y+ = + + + = + + +

(b) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2i ( i ) iz z x y x y x x y y− = + − + = − + −

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3.3 Multiplication of Complex Numbers

(a) ( )1 1 1 1 1i ikz k x y kx ky= + = + k, ∈

2

)

(b) 21 2 1 1 2 2 1 2 1 2 1 2 1( i )( i ) i i iz z x y x y x x y x x y y y× = + + = + + +

1 2 1 2 1 2 2 1( ) i(x x y y x y x y= − + + 3.4 Division of Complex Numbers

This is done by realising the denominator.

21 1 1 2 2 1 2 1 2 1 2 2 1

2 222 2 2 2 2 2 2

1 2 1 2 2 1 1 22 2

2 2

1 2 1 2 2 1 1 22 2 2 2

2 2 2 2

i i i i ii i i

( ) i(

i

z x y x y x x y y x y x yz x y x y x y

)x x y y x y x yx y

x x y y x y x yx y x y

+ − − − += ⋅ =

+ − −+ + −

=+

⎛ ⎞+ −= + ⎜ ⎟

+ +⎝ ⎠

Example 2 Express the following in the form i , where ,x y x y+ ∈R .

(a) (2 + 3i) − (−1 + 2i) (b) (2 + 4i)(1 – i) (c) i34i105

++

Solution (a) (2 + 3i) − (−1 + 2i) = 2 + 3i + 1 – 2i = 3 + i (b) (2 + 4i)(1 – i) = 2 – 2i + 4i – 4i2 = 2 + 2i + 4 = 6 + 2i

5 10i 5 10i 4 3i 20 15i 40i 30 50 25i. 24 3i 4 3i 4 3i 16 9 25+ + − − + + +(c) i= = = = ++ + − +

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Example 3 Solve the simultaneous equations i)i1(i3and0i2)i1( =++=+− wzwz .

Solution

(1 i) 2i 0 ----- (1)z w− + = ; 3i (1 i) i ----- (2)z w+ + =

From (1), we have 2i1 i

wz −=

−. Substituting into (2), we get

( )2

2i3i (1 i) i 1 i

6 (1 i) i1 i6 (1 i ) i 1 i

8 i 11 1 i8 8

w w

w w

w ww

w

−⎛ ⎞ + + =⎜ ⎟−⎝ ⎠

+ + =−+ − = −

= +

= +

Example 4 Find the square root of . i43 + Solution Let i 3x y+ = + 4i . Then ( )2 2 2i 3 4i 2i 3 4ix y x y xy+ = + ⇒ − + = + . Comparing the real and imaginary parts on both sides, we have:

2 2 3 ----- (1)x y− = ; 2 4 ----- (2)xy =

From (2), we have 2 ----- (3)xy

= .

Substituting (3) into (1), we get 2

2

4 2

4 2

2 2

2 2

4 3

4 33 4 0

( 4)( 1) 04 (rej.) or 1

yy

y yy yy y

y y

− =

− =

+ − =

+ − =

= − =

Therefore,

( )1 1 12i i 1 i 18 8 4

1 i 1 i 4z

⎛ ⎞− + −⎜ ⎟⎝ ⎠= = =− −

Thus, Hence, . 1y = ± 2x = ± The square root of 3 + 4i are 2 + i or − (2 + i).

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§4 Complex Conjugates

If z = x + iy, then the conjugate of z, denoted by z*, is defined by z* = x – iy.

yxz i+= and are called a conjugate pair. yxz i* −=

Some important results:

The product of any complex number and its conjugate is real. 2 2 2 2 2* ( i )( i ) i i izz x y x y x y yx xy x y= + − = − + − = +

z + z* = 2Re(z) z – z* = i 2Im(z) (z*)* = z (kz)* = kz*, where k ∈ (z1 ± z2)* = z1* ± z2*

(z1 z2)* = z1*. z2* (zn)* = (z*)n, where n ∈

*

1 1

2 2

*

*z zz z

⎛ ⎞=⎜ ⎟

⎝ ⎠

Can you verify the above results?

Example 5 If and , find i21 +=z i32 −=z ( )1 22 *z z+ and ( )1 2 *z z .

Solution ( ) ( ) ( )1 2 1 22 * 2 * * 2 2 i 3 i 4 2i 3 i 7 iz z z z+ = + = − + + = − + + = −

( ) ( )( ) 21 2 1 2* * * 2 i 3 i 6 2i 3i i 7z z z z= = − + = + − − = i−

Example 6 If , find zz*. 4 9iz = +

Solution

( )( )* 4 9i 4 9i 16 81 97zz = + − = + =

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Example 7

Let w = 1 + i and t ∈ . Given that z = x + iy and *wz wtt

= + , show that z lies on the

curve . 2 2 4x y− =

Solution

( ) ( )1 i *i 1 i

1 ii

1 ii

1 1i

x y tt

t tt

t tt t

t tt t

++ = + +

−= + +

= + + −

⎛ ⎞ ⎛

From *wz wtt

= + , we get

⎞= + + −⎜ ⎟ ⎜

⎝ ⎠ ⎝⎟⎠

Comparing real and imaginary parts,

1x tt

= + and 1y tt

= − ------------ (*)

Substitute (*) into LHS of 2 2 4x y− = , we get

2 22 2 2 2

2 2

1 1 1 1LHS 2 2 4 RHSx y t t t tt t t t

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = + − − = + + − − + = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Thus, z lies on the curve . (Shown) 2 2 4x y− =

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§5 Complex Numbers in GC This section aims to equip you with some basics of using GC in complex numbers, however you are also strongly encouraged to read up any GC guidebook to acquire the basic skills required to utilise a GC and to explore its various functions. Note that you may use the GC to help you in your calculations involving complex numbers, unless it is stated that the problem given needs to be solved without the use a calculator, or if the exact solution is required. Getting Started

1. Press to display mode settings. Scroll down to select “a + b i ” so that you can obtain complex number solutions in Cartesian form.

It is also recommended that you use radian mode for calculations involving complex numbers.

2. To enter the complex number i, press . 3. Other operations or functions for complex numbers can be

found in the Math CPX menu, which contains standard operations involving complex numbers.

Press to get to the CPX menu. A brief description of the 7 operations or functions in the menu:

Operation: Form : Explanation: conj(complex number z) Returns the complex conjugate z* of z. 1: conj( real(complex number z) Returns the real part, x, of z = x + yi = reiθ. 2: real( imag(complex number z) Returns the complex part, y of z = x + yi = reiθ. 3: imag( angle(complex number z) Returns the principal argument, θ, of z = x + yi

= reiθ. 4. angle(

abs(complex number z) Returns the modulus, r, of z = x + yi = reiθ. 5: abs( Complex number z ►Rect Displays z in Cartesian form, z = x + yi. 6: ►Rect Complex number z ►Polar Displays z in Exponential form, z = reiθ. 7: ►Polar

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Example 8 Given that z = 7 17i1 i

− ++

, find (i) z (ii) +2 1*

zz

Solution

Step 1: Store z as 7 17i1 i

− ++

in the GC.

Use to store the complex number as z. After pressing “Enter”, GC will simply express the number in Cartesian form. [Note that you should have changed the mode to “a + bi “ form.]

Step 2: Find z .

Using to call out z, we obtain 3 + 2i as a square root of z. [To check your answer, you may want to square 3 + 2i to see if you can get back z.]

Step 3: Find +2 1*

zz

.

Using conj(z) from MATH CPX menu to represent z*,

we can see that +2 1*

zz

= −118.97 + 120.07i.

You will need to scroll to the right to obtain the numbers.

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§6 Complex Roots of Equations 6.1 Fundamental Theorem of Algebra

An equation of degree n has exactly n roots (real or complex). 6.2 Quadratic Equation

Given a quadratic equation 2 0ax bx c+ + = , where a, b, c are real constants, we get the following cases for the roots:

Value of D = b2 – 4ac Type of Roots D > 0 real and distinct D = 0 real and equal D < 0 complex and occurs in conjugate pair

Example 9 Solve . 0522 =++ xx

Solution Since the coefficients of are real and , where 0,1,2nx n = 2 4 16b ac 0− = − < , we will expect a pair of complex conjugate roots.

22 2 2 4(1)(5) 2 162 5 0 1 2i or 1 2i

2 2x x x

− ± − − ± −+ + = ⇒ = = = − + − −

6.3 Polynomial Equation of Degree n

In general, if z is a complex root of where , then

0.... 012

22

21

1 =++++++ −−

−− axaxaxaxaxa n

nn

nn

n

, for 0, 1, 2, ..., 1, n n−ia i∈ =R z∗ is also a complex root of the equation.

In other words, complex roots of polynomial equations with real coefficients occur in conjugate pairs.

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Example 10 If is a root of , find the other roots. ( i1− ) 0223 =+− xx

Solution Since all the coefficients are real numbers, complex roots must occur in conjugate pairs. Hence the other complex root is ( )1 i+ . The third root must be a real number, a.

By trial and error, we have 3 2 3 2f ( 1) ( 1) ( 1) 2 1 1 2 0, where f ( ) 2x x x− = − − − + = − − + = = − + . Therefore, is the third root. 1a = −Hence the other roots are . 1 and 1 ix x= − = +

A cubic equation with real coefficients has either 3 real roots, or 1 real root and a pair of conjugate complex roots.

Example 11 By completing the square, solve the equation . Explain why the solutions are not a conjugate pair.

0i8)i24(2 =−−+ zz

Solution

From Example 4, we have found that ( )3 4i 2 i+ = ± + .

Hence, ( )2 i 3 4i 2 i 2i or 4z z z+ − = ± + = ± + ⇒ = = − .

( ) ( )( ) ( )( )

2

2 2

2 2

2

(4 2i) 8i 0

2 i 8i 2 i 0

2 i 8i 2 i

2 i 8i 4 4i 13 4i

z z

z

z

z

+ − − =

+ − − − − =

+ − = + −

+ − = + − −

= +

In this example, the complex root does not occur in conjugate pairs since not all the coefficients are real.

Note: The graphic calculator is unable to solve polynomial equations with complex

coefficients.

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§7 Geometrical Representation of Complex Numbers 7.1 The Argand Diagram Imaginary Axis

Im (z)

Real AxisRe (z) O

y

x

P

A complex number can be represented by a vector , where P is the point with Cartesian coordinates

on the x-y plane.

yxz i+=

OP

( yx, )

The x-y plane is called the Argand diagram where the horizontal axis is known as the real axis, denoted by Re(z) and the vertical axis the imaginary axis, denoted by Im(z). Example 12

Represent the following complex numbers in an Argand diagram:

, , , 2 3ia = + 2 3ib = − + 2 3ic = − 2d = , 3ie = , 3f = , 2ig = −

D

AB

Im (z)

Re (z) O

E

F

G C

Question: How are z and z* related on an Argand Diagram?

Answer: They are reflections about the real axis.

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Techniques and operations used in coordinate geometry and vectors can be applied to complex numbers. Addition and subtraction of complex numbers correspond to the parallelogram law of vector addition and subtraction.

Let z1 = a + ib be represented by P1

z2 = c + id be represented by P2. z = z1 + z2 = (a + c) + i (b + d) be represented by P.

In terms of vectors,

1OP OP= +

1 2 2PP OP O=

Re (z) O

z2

z1

Im (z)

1 2z z+

P1

P2

P

z2

z1

Re (z)

Im (z)

2OP

z = z2 − z1 = (c – a) + i (d – b) z2 1z−

P1

P2

In terms of vectors,

1P− O

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7.2 Modulus and Argument

Without loss of generality, let P represent the complex number . We have: i , where ,z x y x y += + ∈R

Modulus of z = length of vector OP = 2 2z r x y= = + Argument of z, arg(z)

= angle between OP and the positive real axis

= 1tan yx

θ −= , where π θ π− < ≤

θ is the principal argument if it lies in the principal range, ie πθπ ≤<− .

Re (z)θ

Py

xO

Im (z)

θ is positive when measured anti-clockwise from the positive real axis and is negative when measured clockwise from the positive real axis.

It is important that you first check the quadrant in which z lies before computing the argument.

Example 13 Find the modulus and argument of each of the following complex numbers:

3, –5, –2i, i3 + and i3 −− Solution

3 3, arg(3) 0

5 5, arg( 5)

2i , arg( 2i)

π

= =

− = − =

− = − =

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Note the following results:

Re (z)θ

z = x + iyy

x O

Im (z)

−θ

z* = x – iy −y

r

r

arg(0) is undefined.

z is real ⇔ arg(z) = 0 or π, and

z is purely imaginary ⇔ ( )2

arg π±=z .

z, z* and –z have the same modulus.

arg(z*) = −arg(z)

zz* = |z|2 (A very useful result.)

§8 Representations of a Complex Number 8.1 Cartesian Form and Modulus–Argument / Polar Form

(1) Cartesian / Algebraic Form: yxz i+= From the Argand diagram, we have

r

Im (z)

Re (z) θ

P y

(2) Modulus–Argument / Polar Form: ( )cos isinz r θ θ= + ,

where 2 2 10 & tan yr x yx

π θ π−= + > − < = ≤ .

Example 14 Express (i) –3 and (ii) in polar form. 5 5i− +

Solution (i)

3 3, arg( 3)3 3(cos isin )

ππ π

− = − =

⇒ − = +

(ii)

(ii) 35 5i 25 25 50 5 2, arg( 5 5i)

4 43 35 5i 5 2 cos isin4 4

π ππ

π π

− + = + = = − + = − =

⎛ ⎞⎛ ⎞ ⎛ ⎞⇒ − + = +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

2 2

cossin

tan

x ry r

z r x yyx

θθ

θ

==

= = +

= x O

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8.2 Euler’s Formula & Exponential Form

From the Maclaurin’s expansion of cos , sin and exθ θ for all real values of x and θ, we have

...!6!4!2

1cos642

+−+−=θθθθ , ...

!5!3sin

53

−+−=θθθθ ,

2 3 4

e 1 ...2! 3! 4!

x x x xx= + + + + +

If we let θi=x , we get

( ) ( ) ( )2 3 4

2 3 4

2 4 3

i i i ie 1 i ...

2! 3! 4!1 1 11 i i ...2! 3! 4!

1 1 1 1 ... i ... cos isin2! 4! 3!

θ θ θ θθ

θ θ θ θ

θ θ θ θ θ

= + + + + +

= + − − + +

⎛ ⎞ ⎛ ⎞= − + + + − + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

θ

Euler’s Formula: ie cos i sinθ θ θ= + In general, ( )ie cos isinr rθ θ θ= + .

Hence, any complex number can be written in the exponential form iez r θ= since z can be expressed in the polar form (cos i sin )r θ θ+ .

Note the following:

1. For iez r θ= , θ must be in radians and πθπ ≤<− . 2. i 2e cos i sin cos sin 1θ θ θ θ θ= + = + 2 =

)

3. i i(2e e nθ π θ+= , for n ∈4. For ,x y∈R , i i ie e .e e e ex y x y x y+ x= = =

( ) ( ) ( ) ( )i i iarg e arg e .e arg e arg e 0x y x y x y y y+ = = + = + =

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Example 15 Express the following in exponential form:

(i) (ii) i11 +=z i32 −=z (iii) 1 i3 i+−

(iv) 3 4(1 i) ( 3 i)+ −

Solution (i) ( )1 11 i arg 1 iz z= + = + = ⇒ = (ii) ( )2 23 i , arg 3 iz z= − = − = ⇒ =

(iii) i4 i

4 6

i6

5i121 i 2e 1 1e e

3 i 2 22e

ππ π

π

π⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞−⎜ ⎟⎝ ⎠

+= = =

−

(iv) 43 3 2i ii3 4 3 4 6 4 34

1 2(1 i) ( 3 i) 2e 2e 32 2e 32 2ez zi

12π π π ππ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞

+ − = = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

* You may use the GC if exact form is not required. 8.3 Conjugate of a Complex Number in Polar Form

If ( )cos isinz r θ θ= + , then ( ) ( )( ) ( )* cos i sin cos i sinz r rθ θ θ= − + − = − θ .

In the exponential form, if iez r θ= , then * iez r θ−= .

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§9 Multiplication and Division of Complex Numbers in Polar Form 9.1 From an Algebraic Point of View

Let 1i1 1 1 1 1(cos i sin ) ez r r θθ θ= + = and 2i

2 2 2 2 2(cos i sin ) ez r r θθ θ= + =

( ) ( )( ) (

( ) ( )

1 2 1 1 1 2 2 2

1 2 1 2 1 2 1 2 1 2

1 2 1 2 1 2

cos i sin cos isin

cos cos sin sin i sin cos cos sin

cos isin

z z r r

r r

r r

θ θ θ θThen,

)θ θ θ θ θ θ θ θ

θ θ θ θ

⋅ = + ⋅ +

⎡ ⎤= − + +⎣ ⎦⎡ ⎤= + + +⎣ ⎦

or ( )1 21 2 1 2 ii i i i1 2 1 2 1 2 1 2e e e e ez z r r r r r r θ θθ θ θ θ +⋅ = ⋅ = ⋅ =

Hence, 1 2 1 2 1 2z z r r z z⋅ = = and ( ) ( ) ( )1 2 1 2 1 2arg arg argz z z z θ θ⋅ = + = +

We can extend the above result to get (a) nn zzzzzzzz ....... 321321 =

(b) ( ) ( ) ( ) ( ) (1 2 3 1 2 3arg .... arg arg arg ..... argn n )z z z z z z z z= + + + +

In particular when , , we have 1 2 3 ... nz z z z= = = = = z

(c) nnz z=

(d) ( )arg arg( )nz n z=

From above, we can see that for n∈ and iez r θ= ,

( ) (i ie e cos i sinnn n n nz r r r n nθ θ )θ θ= = = + --- using Euler’s Formula

ie. ( ) (cos isin cos isinn nr r n )nθ θ θ⎡ ⎤+ = +⎣ ⎦ θ

)

In fact, De Moivre’s Theorem states that

For all real values of n, ( ) (cos isin cos isinnn nz r r n nθ θ θ⎡ ⎤= + = +⎣ ⎦ θ

Also,

( )( )

( ) (( )

)

( ) ( )

1 1 1 1

2 2 2 2

2 21 1 1

2 2 2 2 2

1 1 2 1 2 1 2 1 2

2 22 2 2

11 2 1 2

2

(cos i sin )(cos i sin )

cos i sin(cos i sin ) .(cos i sin ) cos i sin

cos cos sin sin i sin cos cos sin

cos sin

cos isin

z rz r

rr

r

r

rr

θ θθ θ

θ θθ θθ θ θ θ

θ θ θ θ θ θ θ θ

θ θ

θ θ θ θ

+=

+

−+=

+ −

⎡ ⎤+ + −⎣ ⎦=+

⎡ ⎤= − + −⎣ ⎦

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or ( )1

1 2

2

ii1 1 1

i2 2 2

e ee

z r rz r r

θθ θ

θ−= =

Hence, 11 1

2 2 2

zz rz r z

= = and 11 2 1 2

2

arg arg( ) arg( )z z zz

θ θ⎛ ⎞

= − = −⎜ ⎟⎝ ⎠

Note that the results in the boxes are very useful when we need to evaluate the exact values of the modulus and arguments of products and quotients of complex numbers.

9.2 From a Geometrical Point of View

We first consider the special case of multiplication of a complex number by i .

Im (z) Let z = 3 + 2i.

iz = i(3 + 2i) = −2 + 3i

i2z = −(3 + 2i) = −3 − 2i = −z

i3z = −i(3 + 2i) = 2 − 3i

i4z = 3 + 2i = z

Using the same scale on both axes on the Argand diagram, we observe that if a point P represents a complex number z, then the point representing iz is obtained by rotating OP 90o anti-clockwise about the origin.

In general, we consider multiplying a complex number 1i

1 1ez r θ= to another complex number 2i

2 2ez r θ= . The result is ( )1 21 2 1 2 ii i i i1 2 1 2 1 2 1 2e e e e ez z r r r r r r θ θθ θ θ θ +⋅ = ⋅ = ⋅ = .

Geometrically, the length of z1 has been scaled by factor r2, and it has been rotated by θ2 degrees anti-clockwise (assuming θ2 is positive).

Questions to Ponder:

1. What happens when θ2 is negative?

2. How would you illustrate

division geometrically?

Re (z) O

z = 3 + 2i iz = −2 + 3i

i2z = −3 − 2i i3z = 2 − 3i

Im (z)

Re (z) O

1i1 1ez r θ=

( )1 2i1 2 1 2ez z r r θ θ+=

1θ

1 2θ θ+

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Example 16 Express the following in polar and exponential form.

(i) 21 i− +

(ii) 23 )i1()3i1( +−− (iii) 3

6

)i1()3i1(

+−−

Solution

(i) ( )22 2 22, arg arg(2) arg 1 i1 i 1 i 1 i 42

3π⎛ ⎞= = = = − − + = −⎜ ⎟− + − + − +⎝ ⎠

Thus, 3i42 3 32 cos i sin 2e

1 i 4 4

π

π π⎛ ⎞−⎜ ⎟⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞= − + − =⎜ ⎟ ⎜ ⎟⎢ ⎥− + ⎝ ⎠ ⎝ ⎠⎣ ⎦

.

( ) ( )3 23 23 2(1 i 3) ( 1 i) 1 i 3 1 i 1 3 1 1 16− − + = − − + = + + = (ii)

( )( )

3 2 3 2

1

arg (1 i 3) ( 1 i) arg(1 i 3) arg( 1 i) 3arg(1 i 3) 2arg( 1 i)

3 tan 3 24 2π ππ−

− − + = − + − + = − + − +

⎛ ⎞= − + − =⎜ ⎟⎝ ⎠

Thus, i

3 2 2(1 i 3) ( 1 i) 16e 16 cos i sin2 2

π π π⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞− − + = = +⎜ ⎟

⎝ ⎠.

(iii) ( )( ) ( )

666 6

3 3 33

1 31 i 3(1 i 3) 2 16 2( 1 i) 1 i 1 1 2

+−−= = = =

− + − + +

( )6

13

6

3

(1 i 3) 17arg 6arg(1 i 3) 3arg( 1 i) 6 tan 3 3( 1 i) 4 4

(1 i 3)principal arg( 1 i) 4

ππ π

π

−⎛ ⎞− ⎛ ⎞= − − − + = − − − = −⎜ ⎟ ⎜ ⎟⎜ ⎟− + ⎝ ⎠⎝ ⎠⎛ ⎞−

= −⎜ ⎟⎜ ⎟− +⎝ ⎠

Thus, 6 i

43

(1 3i ) 16 2e 16 2 cos isin( 1 i) 4 4

π π π⎛ ⎞−⎜ ⎟⎝ ⎠ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = − + −⎜ ⎟ ⎜ ⎟⎜ ⎟− + ⎝ ⎠ ⎝ ⎠⎝ ⎠

− .

* You may use the GC if exact form is not required.

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Example 17

The complex number q is given by i

i

e1 e

qθ

θ=−

, where πθ 20 << . In either order,

(i) find the real part of q (ii) show that the imaginary part of q is 2

cot21 θ .

Solution

i ii 2 2

i i ii2 2 2

cos i sin cos i sine e e 2 2 2 2.1 e e e e 2i sincos i sin cos i sin

22 2 2 2

i cos sin 1 12 2 i cot2 2 22sin

2

qθ θθ

θ θ θθ

θ θ θ θ

θθ θ θ θ

θ θθ

θ

−

− −

+ += = = =

− −− ⎛ ⎞ ⎛ ⎞− −+ − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−= = − +

Thus, 1Re( )2

q = − , and 1Im( ) cot2 2

q θ= .

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§10 nth Roots of a Complex Number

Suppose we have a fixed number (real or complex), p.

12p p= = square root of p

133 p p= = cube root of p

144 p p= = _______________

155 p p= = __________________

1nn p p= = _______________

Suppose we need to solve the equation nz p= , where p is any fixed number (real or complex). How many roots will we obtain for the following cases?

: No of roots = 2 2z p= : No of roots = 3 3z p= : No of roots = n nz p= To know the number of roots, we look at the highest degree of z in the equation. In this section, we will be looking at obtaining the roots (or solutions) when solving

equations of the form . nz p= Suppose . We have ip x y= + inz p x y= = + , where ir x y+ and ( )arg ix yθ = + . =

Step 1: Convert to exponential form.

ii enz x y r θ= + = , where ir x y= + and ( )arg ix yθ = + Step 2: Add 2kπ to the argument.

( )i 2e knz r π θ+= , k∈

Step 3: Raise both sides by power 1n

.

( )21 1i 2 2e cos i sin

kn n n k kz r r

n n

π θ π θ π+ θ⎡ + +⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎤

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Step 4: Obtain the roots by substituting suitable integer values of k.

The solutions to are known as the nnz = p th roots of the complex number p.

Note: When p = 1, we obtain the special case 1nz = .

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The solutions are called the nth roots of unity. Questions to Ponder:

1. In Step 1, why do we want to convert to exponential form?

Ans: Notice that in Step 3, we need to raise both sides to power of 1n

. It is more

efficient to work in exponential form when dealing with powers of functions.

2. In Step 2, why do we add 2kπ? Ans: The argument of a complex number is not unique.

In an Argand diagram, we can

obtain the same complex number by rotating the argument by 2π, 4π, 6π, etc. (integer multiples of 2π) either clockwise or anti-clockwise about the origin.

In general, we can add/subtract

integer multiples of 2π to the argument of a complex number without changing its final value.

Re (z)θ

Py

x

Im (z)

+ 2π − 2π

3. In Step 4, how many values of k do we need to obtain all the roots?

Ans: The number of values of k to take, depends on the number of unique roots of the equation. In general, if the equation has n unique roots, we will take n consecutive integer values of k. The choice of values of k determines whether we obtain roots with arguments within the principle range.

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Example 18 Find the cube roots of unity of the equation 3 1 0z − = . Display the roots on an Argand diagram.

Solution

3 3 i21 e kz z π= ⇒ = k∈, . 2i

3ek

zπ

= , k = 0, ±1 Taking k = −1, 0, 1 to find the unique solutions, we get

2i3 2 2 1e cos i sin i

3 3 2z

π π π⎛ ⎞−⎜ ⎟⎝ ⎠ ⎛ ⎞ ⎛ ⎞= = − + − = − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠3

2k = −1,

k = 0, z = 1 2i3 2 2 1e cos i sin i

3 3 2z

π π π= = + = − +

32

k = 1,

Re (z)

Im (z)

1 3i2 2

− +

1 3i2 2

− −

1

23π

23π

−

The roots of the equation lie along the circumference of a circle of radius 1.

Note:

• While we could have chosen any 3 consecutive values of k (0, 1, 2 or −3,−2, −1), we often choose values of k which result in principal arguments for the roots. If the argument is not in the principal range, we can easily convert to principal argument.

4i3 4 4 2 2 1e cos i sin cos isin i

3 3 3 3 2z

π π π π π⎛ ⎞ ⎛ ⎞= = + = − + − = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

32

Example: k =2,

This is equivalent to the case when we take k = −1.

• The angle between each root on the Argand diagram is equal.

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Example 19 5 5 i

2 3iz +=

+Find, in exact form, all complex numbers z such that . Display the roots on

an Argand diagram. Solution

21i 2 i5 4 5 2055 i 1 i 2e 2 e

2 3i

kkz z

π π ππ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+

= = − = ⇒ =+

k, ∈

We take k = −2, −1, 0, 1, 2 (or any 5 consecutive integers) to get unique solutions.

( )2 2 171 11 i i5 20 2010 105 17 172 e 2 e 2 cos isin

20 20z

π π π π π−⎛ ⎞ ⎛ ⎞−⎜ ⎟ −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎡ ⎤⎛ ⎞ ⎛ ⎞= = = − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦k = −2,

( )2 1 91 11 i i5 20 2010 105 9 92 e 2 e 2 cos isin20 20

zπ π π π π

−⎛ ⎞ ⎛ ⎞−⎜ ⎟ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎡ ⎤⎛ ⎞ ⎛ ⎞= = = − +⎜ ⎟ ⎜ ⎟k = −1, −⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

( )2 0 1 11 i i5 20 2010 1052 e 2 e 2 cos isin20 20

zπ π π π π

⎛ ⎞ ⎛ ⎞−⎜ ⎟ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎡ ⎤⎛ ⎞ ⎛ ⎞= = = − +⎜ ⎟ ⎜ ⎟k = 0, −⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

( )2 1 71 11 i i5 20 2010 105 7 72 e 2 e 2 cos isin20 20

zπ π π π π

⎛ ⎞ ⎛ ⎞−⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎡ ⎤⎛ ⎞ ⎛ ⎞= = = +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

k = 1,

( )2 2 31 11 i i5 20 410 105 3 32 e 2 e 2 cos isin4 4

zπ π π π π

⎛ ⎞ ⎛ ⎞−⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎡ ⎤⎛ ⎞ ⎛ ⎞= = = +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

k = 2,

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Example 20

Show that the roots of the equation ( )5 51 0z z+ − =1 1 i cot2 5

π⎛ ⎞− ±⎜ ⎟⎝ ⎠

1 21 i cot2 5

π⎛ ⎞− ±⎜ ⎟⎝ ⎠

are and .

Solution

( ) ( )5

5 55 5 11 0 1 zz z z zz+⎛ ⎞+ − = ⇒ + = ⇒ =⎜ ⎟

⎝ ⎠1

2i5 5 i2 51 e ek

kπ

πω ω ω= ⇒ = ⇒ =1z

zω += k∈, . , then Let

1zz

ω +=

11

zω

=−

11

zω

=−

. Hence, are the roots of the equation. , we have From

1 0 or positive and negative multiples of 5kω ≠ ⇒ ≠ . However,

Solving for z,

i5

2i ii i5 55 5

1 1 1ecos i sin cos sine 1 e e e 5 5 5

1 1 1 1 1cos sin cot i cot5 5 2i 5 2 2 2 52isin

5

k

k kk kz k k k i

k k k ki k

π

π ππ π

5kπ π π

π π π ππ

⎛ ⎞−⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟⎝ ⎠

= = = ⋅⎛ ⎞ + − +− −⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= − ⋅ = − = − −⎜ ⎟⎝ ⎠

π

When k = −2, 1 1 2i cot2 2 5

z π= − +

1 1 i cot2 2 5

z π= − + ; When k = −1,

1 1 i cot2 2 5

z π= − −

1 1 2i cot2 2 5

z π= − −When k = 1, ; When k = 2,

1 1 i cot2 5

π⎛ ⎞⎜ ⎟⎝ ⎠

1 1 i cot2 5

2π⎛− ±⎜⎝ ⎠

⎞⎟Hence, the roots of the equation are and . (Shown) − ±

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§11 Loci in the Complex Plane

( ),x yyxz i+=In the Argand diagram, is represented by the point . If the values of x and y vary according to some given condition, the set of points in the Argand diagram will describe some line or curve. This line or curve is called the locus of z.

11.1 Circle with Centre A (represented by a) and Radius, r

The locus of a circle on an Argand diagram is written as z a r− = , where z is a variable complex number, a is a fixed complex number, and r is a fixed real number.

O

Im (z)

Re (z)

P

A

Why does z a r− = describe a circle?

Let P be a variable point and A be the fixed point. Let and OA represent z and a respectively. OPSo represents (i.e. AP AP OP OA= −az − ), and

distance between points and AP z a P A= − = .

AP z a r= − = ∈RSince , where r is fixed, then P

must move in a circle with centre at the fixed point A and radius r.

The locus of z expressed in the form: is a circle raz =−with centre at the fixed point A and radius r.

Let and yxz i+= βα i+=a , we get

( ) ( ) ( ) ( )2 2 2 2 2( ) ( )ix y r x y r x yα β α β α β− + − = ⇒ − + − = ⇒ − + − = r

( ),α β with radius r. which is the Cartesian equation of a circle centred at

Example 21 Describe the locus of z completely for the following:

3i1 =−−z 12i2 =−− z (b) (a)

Solution ( )1 i 1 i 3z z− − = − + =(a) ⇒ ( )The locus of is a circle centered at 1,1 with radius 3.z

i i i i2 i 2 2 1 2 1 2 1 1 12 2 2 2

z z z z z⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − = − − + = − − − = − − = ⇒ − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

12

(b)

1 1The locus of is a circle centered at 1, with radius .2 2

z ⎛ ⎞⇒ − ⎜ ⎟⎝ ⎠

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11.2 Perpendicular Bisector of the Line Joining 2 Points Represented by a and b

The locus of a perpendicular bisector on an Argand diagram is written as z a z b− = − , where z is a variable complex number, and a and b are the fixed complex numbers.

B

O

Im (z)

Re (z)

A P

Why does z a z b− = − describe a perpendicular bisector?

Let P be a variable point and A and B be the fixed points. Let and represent z, a, b respectively. ,OP OA OB

AP BPSo represents and az − bz − represents .

Hence we have AP BP z a z b= ⇔ − = −

ie. the distance between P and A and that of P and B are the same. Hence, P must move along a perpendicular bisector of line joining A and B.

The locus of z expressed in the form bzaz −=−

is the perpendicular bisector of the line joining the fixed points A and B.

Example 22 Describe the locus of the point P representing z when

1ii

=−+

zzi3i3 ++=−− zz . (a) ; (b)

Solution ( ) ( )3 i 3 i 3 i 3 iz z z z− − = + + ⇒ − + = − − −(a)

The locus of z is a perpendicular bisector of the line joining (3,1) and (−3,−1). ⇒

( ) ( )i 1 i i i i ii

z z z z z zz

+= ⇒ + = − ⇒ − − = − − = −

− (b)

The locus of z is a perpendicular bisector of the line joining (0, −1) and (0, 1). ⇒

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11.3 Half–line from Point Represented by a (excluding the point), Making an Angle of α with the Positive Real Axis

A

O

Im (z)

Re (z)

α

P

The locus of a half-line on an Argand diagram is written as ( ) α=− azarg , where z is a variable complex number and a is the fixed complex number, and π α π− < ≤ .

Why arg (z – a) = α describes a half-line?

Let P be a variable point and A be the fixed point.

OALet OP and represent z and a respectively. So, represents . AP az −

( ) α=− azarg AP is the angle that the vector Then makes with the positive real axis. Thus P moves along the half-line that makes an angle α with the positive real axis.

The locus of z expressed in the form ( ) α=− azarg is the half–line from the fixed point A (excluding the point A) that makes an angle α with the positive real axis.

Question to Ponder: Why must we exclude point A?

Example 23

( )3

2i32arg π−=++z( )

3arg π

=zDescribe the locus of z if (a) ; (b)

4i21i2arg π=⎟

⎠⎞

⎜⎝⎛

+−−z( ) π=−1iarg z ; (d) (c)

Solution

( )3

arg π=z ⇒ The locus of z is a half-line from the origin, making an angle of

(a)

3π with the positive real axis.

( ) ( )( ) 2arg 2 3i arg 2 3i3

z z π+ + = − − − = − (b)

⇒ The locus of z is a half-line from (−2, −3), making an angle of 23π

−

with the positive real axis.

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( ) ( )( ) ( ) ( )1arg i 1 arg i arg i i arg i arg ii

z z z z π⎛ ⎞⎛ ⎞− = − = + = + +(c) =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

( )arg i2 2

z π ππ⇒ + = − =

2π The locus of z is a half-line from (0, −1), making an angle of with ⇒

the positive real axis.

( ) ( )2 iarg arg 2 i arg 1 2i1 2i 4

z z π− −⎛ ⎞ = − − − + =⎜ ⎟+⎝ ⎠ (d)

( ) ( )( )1 1arg 2 i tan 2 arg 2 i tan 24 4

z zπ π− −− − = ⇒ − + = + ⇒ −

1tan 24π −+ The locus of z is a half-line from (2, 1), making an angle of ⇒

with the positive real axis. Important Note:

1. We are able to describe the loci of a circle, perpendicular bisector and half-line if they fall into one of the following categories:

( ); ; arg ,z a r z a z b z a θ− = − = − − = where a and b are fixed complex numbers and r, θ are fixed real numbers.

2. The coefficient of z is or must be 1 in the standard equation of the loci.

3. If the equation given is not one of the above, we will either try to change it to one of

the above by algebraic manipulation, or try to obtain the Cartesian equation and then describe the locus. A useful result to use is 2 2 2. *z z z x y= = + .

Example 24

1 2 1z z− = +Find the locus of .

Solution Let . Then, iz x y= +

( )

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

( ) ( )

2

2

2 22

2 22

2 2

2 22 2

1 2 1

i 1 2 i 1

1 2 1 2

1 2 1 2

3 3 6 0

2 0, i.e. 1 0 1

z z

x y x y

x y x y

x y x y

x y x

x y x x y

− = +

+ − = + +

− + = + +

− + = + +

+ + =

+ + = + + − =

Thus, locus is a circle centred at (−1, 0) with radius 1.

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11.4 Problems Involving Inequalities and Maximum/Minimum

Example 25 2i33 =+−z 1+zGiven that , find the greatest and least values of .

Solution

( )3 3i 3 3i 2z z− + = − − =

( )The locus is a circle centered at 3, 3 with radius 2.⇒ −

( ) ( )1 distance between and 1,0z z− − = −P

From the Argand diagram, the least value of 1 9 16z PQ+ = = + − =2 3

O 3

-3 R

Re (z)

Im (z)

( )1, 0P −

Q

,

1 9 16 2z PR 7+ = = + + = . the greatest value of

Example 26

( )3

3arg π=+z , find the least value of |z|. If

Solution

The least value of |z| is the distance between P and the origin.

3 3Least value of 3sin3 2

z OP π= = =

O - 3 3π

Re (z)

Im (z)

P

Example 27

Shade the region represented on an Argand diagram by

(a) 2i ≤−z (b) ( )6

i2arg π2

arg4

ππ<< z<++< z (c) i22 −−≤ zz0 and

Solution

( )i iz z− = − =

1

O

-1

3

Re (z)

Im (z) (a) 2 ⇒ The locus of z is a circle centred at (0, 1) with radius 2.

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( )6

i2arg0 π<++< z(b)

( )( )arg 2 i6

z π− − − =

The locus of z is a half-line from ( , ), making an angle of ⇒

6π with the positive real axis.

2arg

4ππ

<< z i22 −−≤ zz(c) and

Re (z) O 2

2

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Example 28 On a single Argand diagram sketch the loci given by

3 4z − = (i)

3 3iz z− − = (ii) Hence, otherwise, find the exact values of all the complex numbers z that satisfy both (i) and (ii).

Solution

(i) 3 4z − = is a circle centred at (3, 0) with radius 4.

3 3iz − − = z (ii) is a perpendicular bisector of the line joining the points (3, 3) and (0, 0).

Locus of (i)

Im( )z

Re( )z 7 -1 3

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Alternatively using Graphic Calculator, Step 1: Find Cartesian equation of circle : ( ) (2 22 2 23 4 16x y y x− + = ⇒ = − − )3

( )21Y 16 3x= − − ( )22Y 16 3x= − − − Key in and or 1Y− Step 2: Find Cartesian equation of the perpendicular bisector:

3 0 3 13 0 3−

= =−

Gradient of line joining (3, 3) and (0, 0) =

1 11

− = −Thus gradient of the perpendicular bisector is and passes through the mid

point of (3, 3) and (0, 0) i.e. 3 0 3 0, 2 2+ +⎛ ⎞

⎜ ⎟⎝ ⎠

.

Equation of the perpendicular bisector:

( )3 3 12 2

3

y x

y x

⎛ ⎞− = − −⎜ ⎟⎝ ⎠

⇒ = − +

Key in 3Y 3x= − + Step 3: Press ‘Graph’. Enter ‘2nd’, ‘trace’ and scroll down to press ‘intersection’ to find the intersection(s) of the circle and the perpendicular bisector.

Note: You may use the GC to help you in your calculations involving complex numbers, unless it is stated that the problem given needs to be solved without the use of a calculator, or if the exact solution is required.

The End

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nj complex numbers lecture notes student edition

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