Ch04 Relations And Regions Of The Complex Plane 312031

4

4A 4B 4C 4D

Rays and lines Circles and ellipses Combination graphs and regions Graphs of other simple curves

Relations and regions of the complex plane aReaS oF STudy

Representations of relations and regions of the complex plane including: • Lines and rays

• Combinations of the above • Regions defined through the above curves • Circles, ellipses and other familiar simple curves eBook plus Digital doc

4a

Rays and lines

10 Quick Questions

introduction We have seen in chapter 1 that relations between real variables x and y can be represented by a set of points on the Cartesian plane. The visual display of a graph gives a better understanding of a relationship between two variables and this can help us greatly in problem solving. Similarly, rules or relationships defined by a complex variable can be better understood when represented on the complex plane. For example, to graph the complex relations {z: Re(z) + 3Im(z) = 6} or {z: | z − 2 + 3i | = 5} would at first appear daunting but, as we will soon see, by using our ‘complex number’ knowledge it is not too difficult and, in fact, is effectively the same as graphing on the Cartesian plane. First, recall the following facts from the chapter on complex numbers. 1. A complex number, z, consists of two parts: z = x + yi where x, y ∈ R x is called the real part of z: x = Re(z). y is called the imaginary part of z: y = Im(z). 2. Complex numbers can be represented on the complex plane, or Im(z) Argand diagram, with the origin 0 + 0i (see figure at right): Consider the point P in the second Argand diagram. This represents the complex number z = x + yi.

162

maths Quest 12 Specialist mathematics for the Casio Classpad

0

Re(z)

Im(z)

3. The modulus or absolute value of z is defined as the distance from the origin to the point P.

z = x + yi P = (x, y)

mod ( z ) = | z | = x 2 + y 2 Arg(z)

0

Re(z)

4. The argument of z, Arg(z), is defined as the angle between OP and the positive real axis. −π < Arg(z) ≤ π

Rays Consider {z: Arg(z) = θ}. This expression defines the set of points which make an angle of θ with the positive real axis. This is represented on the complex plane by a ray which starts at the origin and makes an angle of θ with the positive real axis: Note: An open circle at the origin shows that 0 + 0i is not a member of the set.

Im(z) {z: Arg(z) = θ}

θ 0

Re(z)

Worked Example 1 − 5π   Sketch  z: Arg( z ) =  on the complex plane. 6  

Think 1

Write

Draw a complex plane.

Im(z) 0

2

− Mark a point which is approximately 5π radians from 6 the positive real axis (3rd quadrant).

Im(z) 0 π − 5— 6

3

Draw a ray from the origin through this point.

Re(z)

Re(z)

Im(z) 0

Re(z)

π − 5— 6

Im(z)

Translation of rays The translation of a complex relation is similar to translations on the Cartesian plane. For example, the relation y + 1 = (x + 2)2 is the same as y = x2 translated −2 units in the x direction and −1 unit in the y direction. Consider the ray defined by {z: Arg(z + a + bi) = θ, a, b ∈ R}.

Assuming a, b > 0 –a

θ

0

Re(z)

θ –b −a − bi {z: Arg (z + a + bi) = θ }

Chapter 4  Relations and regions of the complex plane

163

This is the same set as {z: Arg(z) = θ} translated −a units in the ‘real’ direction and −b units in the ‘imaginary’ direction. Note: The ray starts at the point (−a − bi), making an angle of θ with the positive side of the horizontal line ing through −b on the imaginary axis. Worked Example 2

π  Show  z: Arg( z + 4 − 2 i ) =  on an Argand diagram. 3  Think 1

Write

π  Sketch z: Arg( z ) =  on a complex plane. 3 

Im(z) π– 3

0 2

Translate the ray −4 units in the real direction and +2 units in the imaginary direction.

Re(z)

Im(z) π– 3

2

π– 3

0

−4

Re(z)

Lines Consider {z: Re(z) = a, a ∈ R}. This defines the set of points whose real part is a (that is, x = a). It is represented on the complex plane by a vertical line ing through the point a + 0i. Now consider {z: Im(z) = b, b ∈ R}. This defines the set of points whose imaginary part is b (that is, y = b). It is represented on the complex plane by a horizontal line ing through the point 0 + bi. Im(z) b a Re(z) 0 Assuming b>0 {z: Im(z) = b}

Worked Example 3

Sketch the graphs of the following on the complex plane. 7

a   {z: Re(z) = 2 }   b   {z: Im(z) = −2} Think a Sketch a vertical line through the point

complex plane.

164

Write 7 2

+ 0i on the

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

a

Im(z) 0

7– 2

Re(z)

Im(z) 0

a Re(z)

Assuming a>0 {z: Re(z) = a}

b Sketch a horizontal line through the point 0 − 2i on

b

Im(z)

the complex plane.

0

Re(z)

−2

Translation of lines

Im(z)

The graph of {z: Re(z + a + bi) = c, where a, b, c ∈ R} is the same as the graph of {z: Re(z) = c} translated −a units in the real direction and −b units in the imaginary direction. Note: Since the graph is a vertical line the translation in the imaginary direction is meaningless.

c−a

−a 0

−b

c Re(z)

−b {z: Re(z + a + bi) = c} Assuming a, b, c > 0 and a > c

Likewise the graph of {z: Im(z + a + bi) = c, where a, b, c ∈ R} is the same as the graph of {z: Im(z) = c} translated −a units in the real direction and −b units in the imaginary direction. Note: Since the graph is a horizontal line the translation in the real direction is meaningless.

Im(z) c −b −a

c−b

0 Re(z) −b {z: Im(z + a + bi) = c} Assuming a, b, c > 0 and b < c

Worked Example 4

Show on an Argand diagram: a   {z: Re(z − 2 + 3i) = 1} b   {z: Im(z − 5i) = −3}. Think a

1

2

b

1

Write

Sketch the graph of {z: Re(z) = 1} on a complex plane.

a

1 Re(z)

0

Translate the line +2 units in the real direction (and −3 units in the imaginary direction).

Sketch the graph of {z: Im(z) = −3} on a complex plane.

Im(z)

Im(z)

2 0 1

b

3 Re(z)

Im(z)

0

Re(z)

−3

Chapter 4  Relations and regions of the complex plane

165

2

Translate the line +5 units in the imaginary direction.

Im(z) 2 0

Re(z)

−3



1. The graph of {z: Arg(z) = θ} is a ray starting at the origin and making an angle of θ with the positive real axis. 2. The graph of {z: Arg(z + a + bi) = θ} is the same as {z: Arg(z) = θ} translated −a units in the real direction and −b units in the imaginary direction. 3. The graph of {z: Re(z + a + bi) = c} is a vertical line ing through the point (c − a) + 0i. 4. The graph of {z: Im(z + a + bi) = c} is a horizontal line ing through the point 0 + (c − b)i.

Exercise

4A

Rays and lines   1   WE 1    Sketch each of the following on separate complex planes.

π  a  z: Arg( z ) =  6 

3π   b  z: Arg( z ) =  4  

−π   c  z: Arg( z ) =  4 

− 2π   d  z: Arg( z ) =  3  

e {z: Arg(z) = π}

−π   f  z: Arg( z ) =  6 

  2   WE2    Show each of the following on separate Argand diagrams.

166

π  a z: Arg( z − 2) =  2 

2π   b  z: Arg( z + 3) =  3 

− 3π   c z: Arg( z − 1) =  4  

d z: Arg(z + 2i) = π

− 5π   e  z: Arg( z − 3i) =  6  

−π   f z: Arg( z + i) =  4 

π  g  z: Arg( z + 2 − i) =  3 

−π   h  z: Arg( z − 3 + 2i) =  8 

i  z: Arg( z − 4 + 3i) = π  2 

− j  z:Arg( z + 2) = π  5 

{

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

}

3π     3   MC  Consider the ray represented by  z: Arg( z ) = . 4  a If the ray is translated −2 units in the imaginary direction it could be described by: 3π   A  z: Arg( z − 2i) =  4 

− 3π   B  z:Arg( z + 2) =  4  

3π   C  z: Arg( z + 2i) =  4 

− 3π   D  z: Arg( z + 2i) =  4  

3π   E  z: Arg( z − 2) =  4  b If the ray is translated −3 units in the real direction and +2 units in the imaginary direction then it can be described by: 3π   A  z: Arg( z − 3 + 2i) =  4 

3π   B  z: Arg( z + 3 + 2i) =  4  

3π   C  z: Arg( z − 3 − 2i) =  4 

− 3π   D  z: Arg( z + 3 − 2i) =  4  

3π   E  z: Arg( z + 3 − 2i) =  4 

−π     4   MC  Which one of the following graphs represents  z: Arg ( z + 1 − i) = ? 4  A Im(z) C Im(z) B Im(z)

−1

D

0 1 Re(z) (1, −1)

(−1, 1) 1

− –π 4

−1 0

E

Im(z) –π 4

0 Re(z) −1 π (−1, −1) − –4 −1

− –π 4

Re(z)

Im(z) (−1, 1)

−–4π

(−1, 1)

0 Re(z)

0 Re(z)

  5   MC  The graph which represents {z: Re(z) = −3} is: B A Im(z) Im(z) 0

3

Re(z)

Im(z)

−3

0 Re(z)

0

3 Re(z)

0 Re(z)

−3

D

C Im(z)

E

Im(z) −3

0 Re(z) −3

  6   WE3a, b    Sketch the graph of each of the following on a separate complex plane. a {z: Re(z) = 5} b {z: Re(z) = −2} c {z: Re(z) = 4}

{

d z: Re( z ) =

5 2

}

Chapter 4  Relations and regions of the complex plane

167

{z: Im(z) = } 3 2

e {z: Im(z) = −4}

f

g {z: Im(z) = 1}

h {z: Im(z) = −3}

  7   WE4a, b    Show each of the following on a separate Argand diagram. a {z: Re(z − 2) = 3} b {z: Re(z − i) = 2} − c {z: Re(z + 3) = 2} d {z: Re(z − 1 + 2i) = 1} e {z: Im(z + 3i) = 2} f {z: Im(z − 2i) = −1} g {z: Im(z + 2) = 3} h {z: Im(z − 2 − 4i) = −2}   8 Sketch each of the following on a separate complex plane. 2π   a  z: Arg( z ) =  3  c {z: Im(z − 2 + i) = 2} e {z: Re(z + 3 + 2i) = 1} 3π   g  z: Arg( z + i) =  8 

4b

b {z: Re(z − 5) = −3}

π  d  z: Arg( z − 1 + 3i) =  2  f {z: Im(z + 5i) = 3} h {z: Im(z − 3i) = 0}

Circles and ellipses

Im(z) r

Circles Since | z | is defined as the distance of a point from the origin on the complex plane, it follows that {z: | z | = r} is the set of all the points on the complex plane that are r units from the origin.

−r

0

r

Re(z)

−r

{z | z | = r} defines a circle of radius r and centre 0 + 0i. Worked Example 5

Illustrate {z: | z | = 5} on an Argand diagram. Think

Write

1

The equation defines a circle of radius 5 and centre 0 + 0i.

2

Sketch this circle on the complex plane.

5

Im(z)

−5 0

5 Re(z)

−5

Translation of circles Consider {z: | z + a + bi | = r, a, b ∈ R}. This defines the same set of points as {z: | z | = r} but translated −a units in the real direction and −b units in the imaginary direction. {z: | z + a + bi | = r} defines a circle of radius r and centre (−a − bi).

Im(z) r −r r −a − bi

−r

0 r Re(z)

Assuming a, b > 0

168

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

WoRked example 6

Sketch each of the following on a separate complex plane. a {z: | z + 1 | = 2} b {z: | z − 2 + 3i | = 3} Think

WRiTe

a This represents a circle of radius 2 and

a

centre −1 + 0i.

Im(z) 2 1 −3 −2 −1 0 1 Re(z) −1 −2

b This represents a circle of radius 3 and centre 2 − 3i.

b Im(z) 0 −1 −3

2 2 − 3i

5 Re(z)

−6

ellipses The Cartesian equation of an ellipse of centre (0, 0) is: x2 y2 + =1 a2 b2 Since z = x + yi where x = Re(z) and y = Im(z), an ellipse can be expressed as a complex relation: (Re( z ))2 (Im( z ))2 + =1 a2 b2 This represents an ellipse which has centre 0 + 0i with semi-major axis length of a and semi-minor axis length of b. The real axis intercepts are ± a and imaginary axis intercepts are ± b.

Im(z) b −a

0

a Re(z)

−b

WoRked example 7 a Express

x2

[3Re(z)]2

+

eBook plus

[2Im(z)]2

= 36 as a Cartesian equation of the form

y2

int-0374

+ = 1. a 2 b2 b Hence sketch the relation on the complex plane. Think a

Tutorial Worked example 7

WRiTe

1

Replace Re(z) with x and Im(z) with y to obtain the Cartesian equation.

2

Expand the brackets.

a (3x)2 + (2y)2 = 36

9x2 + 4y2 = 36

Chapter 4

Relations and regions of the complex plane

169

3

x2 y2 + =1 4 9

Divide both sides by 36 to obtain the simplest form of an ellipse.

x2 y2 + =1 2 2 32 b The relation is an ellipse with centre 0 + 0i, real axis

b

intercepts of ±2 and imaginary axis intercepts of ±3.

Im(z) 3 0

−2

2 Re(z)

−3

y

An ellipse can also be described as the set of points, P, in a plane where the sum of the distances from each point P to two fixed points A and B is a constant. That is, PA + PB = C or P = {z: | z − A | + | z − B | = C} where A and B are two points in the complex plane and C is a real constant.

P

b −a

0

A

B

a

x

−b

Worked Example 8

Show that the locus described by | z − 2 | + | z + 2 | = 8 is an ellipse, and find its Cartesian equation. Think 1

Let z = x + yi and then express | z − 2 | and | z + 2 | in of x and y.

Write

Let z = x + yi | z − 2 | = ( x − 2)2 + y 2 | z + 2 | = ( x + 2)2 + y 2 ( x − 2)2 + y 2 + ( x + 2)2 + y 2 = 8

2

Rewrite the equation in of x and y.

3

Subtract one square root term from both sides.

( x − 2)2 + y 2 = 8 − ( x + 2)2 + y 2

4

Square both sides.

( x − 2)2 + y 2 = 64 − 16 ( x + 2)2 + y 2 + ( x + 2)2 + y 2

5

Expand and simplify.

x 2 − 4 x + 4 + y 2 = 64 − 16 ( x + 2)2 + y 2 + x 2 + 4 x + 4 + y 2 − 8x

− 64 = − 16 ( x + 2)2 + y 2 x + 8 = 2 ( x + 2)2 + y 2

170

6

Square both sides.

7

Expand and simplify.

(x + 8)2 = 4[(x + 2)2 + y2] x2 + 16x + 64 = 4(x2 + 4x + 4 + y2) = 4x2 + 16x + 16 + 4y2 48 = 3x2 + 4y2

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

8

Divide both sides by 48.

9

Simplify.

10

Describe the relation found.

3x 2 4 y 2 + =1 48 48 x2 y2 + =1 16 12 This is an ellipse whose centre is (0, 0), whose semi-major axis is of length 4 and whose semi-minor axis is of length 12 .

Translation of ellipses ( x − h) 2 ( y − k ) 2 + = 1 is an a2 b2 ellipse with centre (h, k). As seen previously the equivalent complex equation is obtained by replacing x with Re(z) and y with Im(z), which gives: The Cartesian equation

[Re( z ) − h]2 a2

+

[Im( z ) − k ]2 b2

Im(z) b+k −a + h

0

−b + k

where h, k ∈ R

=1

h + ki

k

h

a + h Re(z)

Assuming h, k > 0 and a > b

This represents an ellipse with centre h + ki, a semi-major axis length of a and semi-minor axis length of b. WoRked example 9

Illustrate the graph of [Re(z) − 1]2 + 4[Im(z) + 2]2 = 4 on an Argand diagram. Think

WRiTe

1

Express the equation in Cartesian form.

(x − 1)2 + 4(y + 2)2 = 4

2

Simplify by dividing both sides by 4.

( x − 1)2 + ( y + 2)2 = 1 4

3

The relation is an ellipse with centre 1 − 2i, semi-major axis of 2 and semi-minor axis of 1.

Im(z) −1−1 0 1 −2 −3

WoRked example 10

1

Let z = x + yi, and express | z − 3 | and | z − 1 | in of x and y.

3 Re(z)

1 − 2i

eBook plus

Sketch and describe the subset of the complex plane defined by {z: | z − 3 | + | z − 1 | = 4}. Think

2

Tutorial

int-0375 Worked example 10

WRiTe

Let z = x + yi | z − 3 | = ( x − 3)2 + y 2 | z − 1 | = ( x − 1)2 + y 2

Chapter 4

Relations and regions of the complex plane

171

( x − 3)2 + y 2 + ( x − 1)2 + y 2 = 4

2

Rewrite the equation in of x and y.

3

Subtract one square root term from both sides.

( x − 3)2 + y 2 = 4 − ( x − 1)2 + y 2

4

Square both sides.

( x − 3)2 + y 2 = 16 − 8 ( x − 1)2 + y 2 + ( x − 1)2 + y 2

5

Expand and simplify.

x 2 − 6 x + 9 + y 2 = 16 − 8 ( x − 1)2 + y 2 + x 2 − 2 x + 1 + y 2 − 4x

− 8 = − 8 ( x − 1)2 + y 2

x + 2 = 2 ( x − 1)2 + y 2 (x + 2)2 = 4[x - 1)2 + y2]

6

Square both sides.

7

Expand and simplify.

8

Complete a perfect square in x.

0 = 3(x2 - 4x) + 4y2 0 = 3[(x - 4x + 4) - 4] + 4y2 0 = 3[(x - 2)2 - 4] + 4y2 0 = 3(x - 2)2 - 12 + 4y2

9

Add 12 to both sides.

3(x - 2)2 + 4y2 = 12

10

Divide both sides by 12.

3( x − 2)2 4 y 2 + =1 12 12

11

Simplify.

( x − 2)2 y 2 + =1 4 3

12

Describe the relation found.

This is an ellipse whose centre is (2, 0), whose semi-major axis is of length 2 and whose semi-minor axis is of length 3.

13

Sketch the relation.

Im(z)

x2 + 4x + 4 = 4(x2 - 2x + 1 + y2) = 4x2 - 8x + 4 + 4y2 0 = 3x2 - 12x + 4y2

√3 0 −√3

1 2 3 4 Re(z)



1. The relation {z: | z + a + bi | = r} defines a circle of radius r and centre −a − bi. [Re( z ) + h]2 [Im( z ) − k] 2 + = 1 is an ellipse with centre h + ki, a2 b2 semi-major axis length of a and semi-minor axis length of b. 3. {z: | z − A | + | z − B | = C} describes an ellipse where A and B are two points in the complex plane and C is a real constant.

2. The complex equation

172

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

Exercise

4b

Circles and ellipses 1   WE 5 

Illustrate each of the following on a separate Argand diagram.

a {z: | z | = 1} d {z: | z | = 10}

b {z: | z | = 4} e {z: | z | = 9}

c {z: | z | = 3} f {z: | z | = 2}

2   MC  The equation of the circle at right is: A | z + 2 + i | = 2 B  | z − 2 − i | = 2 C | z − 2 + i | = 2 D  | z − 2 − i | = 4 E | z + i + 2 | = 4

Im(z) 3 2 1 0

2+i 1 2 3 4 Re(z)

−1

3   MC  The equation of a circle with centre −3 + 2i and radius 4 is: A | z − 3 + 2i | = 16 B | z − 3 − 2i | = 4 C | z + 3 + 2i | = 4 D | z + 3 − 2i | = 4 E | z + 3 − 2i | = 16 4   WE 6  Sketch the graph of each of the following on separate complex planes. a {z: | z − 3 | = 1} b {z: | z + 2 | = 2} c {z: | z − 4i | = 3} d {z: | z + i | = 4} e {z: | z − 2 − i | = 3} f {z: | z + 3 − 2i | = 1} g {z: | z + 4 + 3i | = 5} h {z: | z − 5 + 5i | = 2} x2 y2 + = 1 is: 5   MC  The complex equation for the ellipse with Cartesian equation 10 5 A [Re(z)]2 + 5[Im(z)]2 = 10 B 2[Re(z)]2 + [Im(z)]2 = 10 2 2 C [Re(z)] + 2[Im(z)] = 10 D [Re(z)]2 + [2Im(z)]2 = 10 E [2Re(z)]2 + [Im(z)]2 = 1 6   MC  The graph which represents (Re(z))2 + 2[Im(z)]2 = 2 on the complex plane is: A Im(z) 2

√2

−1 0 1 Re(z)

Im(z) 1

−1 0 1 Re(z)

−2

−√2

−2

D

C

B Im(z) (2, 1)

E

Im(z) 2

0

2 Re(z)

Im(z) 1

0

−√2

−1

√2 Re(z)

−√2 −1

0

√2 Re(z)

−2

x2 y2 + = 1. a2 b2 b [Re(z)]2 + [3Im(z)]2 = 9 d [Re(z)]2 + 4[Im(z)]2 = 16 f 9[Re(z)]2 + 4[Im(z)]2 = 1

7   WE 7a  Express each of the following as Cartesian equations in the form a [2Re(z)]2 + [Im(z)]2 = 4 c 3[Re(z)]2 + 4[Im(z)]2 = 12 e 4[Re(z)]2 + [Im(z)]2 = 1

8   WE 7b  Sketch each of the relations in question 7 on separate complex planes. 9   WE 8  Show that the locus described by each of the following is an ellipse, and find its Cartesian equation. a | z + 1 | + | z - 1 | = 4 b | z + 3 | + | z - 3 | = 10 c | z + 2i | + | z - 2i | = 6 d | z + 4i | + | z - 4i | = 10

Chapter 4  Relations and regions of the complex plane

173

10 We9 Illustrate each of the following on separate Argand diagrams. b 4[Re(z) − 3]2 + 16[Im(z) − 1]2 = 16 a [Re(z) − 2]2 + 9[Im(z) + 1]2 = 9 2 2 c [Re( z ) + 2] + [Im( z ) −3] = 1 25 9

e 2[Re(z) + 3]2 + 6[Im(z) − 2]2 = 6 g 9[Re(z) − 2]2 + 16[Im(z) − 3]2 = 1

2

2 d [Re( z ) − 4] + [Im( z ) + 3] = 1 16 36 f 3[Re(z) + 3]2 + 5[Im(z) + 2]2 = 15 h 4[Re(z) + 1]2 + 25[Im(z) − 1]2 = 1

11 We10 Sketch and describe the subset of the complex plane defined by each of the following. a {z: | z + 3 | + | z + 1 | = 4} b {z: | z − 3 | + | z + 1 | = 6} c {z: | z − 2 | + | z − 4 | = 10} d {z: | z + 2i | + | z + 4i | = 10} e {z: | z − 3i | + | z − i | = 4} f {z: | z + 3i | + | z − i | = 6} 12 Show that the locus described by | z + 3 | − | z − 3 | = 4 is a hyperbola. State its Cartesian equation.

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WorkSHEET 4.1

4C eBook plus

Combination graphs and regions Combination graphs Clearly it is feasible to represent more than one relationship on the complex plane. The representation of more than one relationship is usually indicated using the ‘union’ symbol ∪. For example, sketch {A} ∪ {B} means sketch both relationship A and relationship B on the same plane.

Interactivity

int-0345 Combination graphs and regions

WoRked example 11

Show {z: | z | = 4} ∪ {z: Re(z − 3) = 1} on an Argand diagram. Think

WRiTe

1

The graph of {z: | z | = 4} is a circle of centre 0 + 0i and radius 4.

2

The graph of {z: Re(z − 3) = 1} is the same as the vertical line ing through the point 1 + 0i but translated +3 units in the real direction.

3

Represent both graphs on the same complex plane.

Im(z) 4 0

–4

4 Re(z)

–4

Regions

π < Arg( z ) ≤ . 3 2 For the boundaries, a solid line is used to represent ≤ or ≥ and a broken line is used to represent < or >. The region required can be shaded in or shaded out as long as it is indicated somewhere on the graph. We will use the convention of shading out because, as will be seen later in this section, it shows the combination of regions more clearly. Regions are described by complex inequations, for example,

174

maths Quest 12 Specialist mathematics for the Casio Classpad

−π

WoRked example 12

eBook plus

2π   −π Represent the region defined by  z: < Arg ( z ) ≤  on a complex plane. 4 3   Think 1

2

3

WRiTe

Use a broken line to represent the boundary −π Arg( z ) = on a complex plane. 4

Tutorial

int-0376 Worked example 12

Im(z) 0

On the same plane use a solid line to represent the 2π boundary Arg( z ) = . 3 Leave the region between the two rays (minor region) unshaded but shade the rest of the plane and indicate that the unshaded region is the region required.

π 2— 3 −–π 4

Re(z)

Im(z) 0

π Region 2— 3 required – –π 4

Re(z)

WoRked example 13

Sketch the graph of each of the following: a {z: Im(z + 4i) > 2} b {z: −1 < Re(z + 2 − 3i) ≤ 5}. Think a

1

2 3

WRiTe

Im(z + 4i) = 2 is a horizontal line ing through the point 0 + 2i translated −4 units in the imaginary direction.

a

On the complex plane draw a broken horizontal line through the point 0 − 2i. Shade the region below the line and indicate the unshaded region as the region required.

Im(z) 0

Re(z)

0

Re(z)

−2 Im(z) Region required

−2 b

1

Re(z + 2 − 3i) = −1 is a vertical line ing through the point −1 + 0i and translated −2 units in the real direction and +3 units in the imaginary direction.

2

On the complex plane draw a broken vertical line ing through the point −3 + 0i and shade the region left of this line.

3

Re(z + 2 − 3i) = 5 is a vertical line ing through the point 5 + 0i and translated −2 units in the real direction and +3 units in the imaginary direction.

b

Im(z)

4

On the same plane draw a solid vertical line ing through the point 3 + 0i and shade the region right of this line.

5

Indicate the unshaded region between the two lines as the region required.

Chapter 4

Region required

−3

0

3 Re(z)

Relations and regions of the complex plane

175

Worked Example 14

Illustrate each of the following on an Argand diagram. a   {z: | z | > 1}    b   {z: 2 ≤ | z + 1 − 4i | < 3} Think a

1

Write

Draw a circle with its centre at the point 0 + 0i and radius 1 on a complex plane. Use a broken line.

a

Im(z) 1 0

−1

1 Re(z)

−1 2

Shade the region inside the circle and indicate the unshaded region as the region required.

Im(z) 1

Region required

0

−1

1 Re(z)

−1 b

1 2

| z + 1 − 4i | = 2 is a circle of centre −1 + 4i and radius 2. Draw this circle with a solid line on the complex plane and shade the region inside the circle.

b

Im(z) 6 −1 + 4i 4 2 1 −3 −1

3 4

5

−1

| z + 1 − 4i | = 3 is a circle with centre + 4i and radius 3. On the same plane draw this circle with a broken line and shade the region outside this circle. Indicate the unshaded region between the two circles as the region required.

Region required

1 Re(z)

Im(z) 7 6

−1 + 4i 4 2 1 −4−3 −1

1 2 Re(z)

Combining regions Two or more regions are combined by representing the area where the regions overlap or intersect. The combining of two or more regions is indicated by the ‘intersection’ symbol, ∩. For example, sketch {A} ∩ {B} requires the intersection of the regions described in A and B to be indicated. Worked Example 15

Sketch the graph for { z:

−π

2

< Arg ( z − 1) ≤ 0} ∩ { z: | z − 1 | > 4 }.

Think 1

176

Write −π

−π

is the same as the ray Arg( z ) = 2 2 translated +1 unit in the real direction. Arg( z − 1) =

2

On a complex plane draw this ray with a broken line.

3

Arg(z − 1) = 0 is the same as the ray Arg z = 0 translated +1 unit in the real direction.

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

Im(z)

0

− –π 2

Re(z)

4

Draw this ray with a solid line on the same plane.

5

Shade out the major region between the two rays.

6

| z − 1 | = 4 is a circle of centre 1 + 0i and radius 4.

7

Draw this circle on the same plane with a broken line.

Im(z) 4 –3

0 1

5 Re(z)

–4 8

Shade out the region inside this circle and indicate the unshaded region remaining as the region required.

Im(z) 4 –3

0 1 –4

5 Re(z) Region required

Note: Wherever practicable, check that the region required is correct by selecting a point in the region and make sure that it satisfies the inequation(s).

1. More than one graph or region can be represented on the complex plane. 2. For regions: (a) The boundaries are represented by:   (i) solid lines where ≤ or ≥ apply (ii) broken lines where < or > apply. (b) Shade the regions not required. 3. Wherever practical, check that the region required is correct by selecting a point in the region and making sure that it satisfies the inequation(s).

Exercise

4c

Combination graphs and regions 1   WE11  Show each of the following on an Argand diagram. a {z: | z | = 2} ∪ {z: Re(z + 2) = −1} b {z: | z | = 3} ∪ {z: Im(z − 3i) = 1}

π  c z: Arg ( z ) =  ∪ {z: Re( z − 2) = − 3} 3  −π   d {z: | z − 2 | = 2} ∪  z: Arg( z + 2i) =  4  e {z: | z + 1 − 3i | = 1} ∪ {z: Re(z − 3 + 2i) = −2} f {z: Re(z + 4) = 2} ∪ {z: Im(z + 10 − 4i) = 0} − 3π   g {z: | z + 3 − 3i | = 2} ∪  z: Arg( z − 2 + i) =  8  

5π   h {z: | z − 2i | = 3} ∪ {z: Re( z + 2) = 1} ∪ z: Arg( z − 4 − 3i) =  9  

Chapter 4  Relations and regions of the complex plane

177

2   MC  The graph shown is described by: 2π   A z: Arg( z ) =  ∪ {z: Im( z ) = − 2} 3  

Im(z)

π  B  z: Arg( z ) =  ∪ {z: Im( z ) = − 2} 3 

π – 3

0 Re(z)

−π

  C  z: Arg( z ) =  ∪ {z: Im( z ) = − 2} 3  

−2

π  D  z: Arg( z ) =  ∪ {z: Im( z − 2) = 0} 3  2π   E  z: Arg ( z ) =  ∪ {z: Im( z − 2i) = 0} 3  3   MC  The region shown is described by: π π  −π  −π ≤ Arg( z ) <  ≤ Arg ( z ) ≤  A  z: B  z: 3 3 2 3    2π   −π ≤ Arg( z ) < C  z:  2 3 

2π   −π < Arg( z ) ≤ D   z:  2 3  

−π

Im(z) –π 3

0 Re(z) Region required

2π   ≤ Arg( z ) ≤ E  z:  2 3   4   MC  The graph which represents the region described by {z: −2 ≤ Im(z + 1 − 2i) < 3} is: A Im(z) C Im(z) B Im(z) 3

5

2 0

D

Region required

Region required

E

Im(z)

0

Re(z)

Im(z) 5

3

Region required

2

0 Re(z)

Re(z)

Region required

3

Region required

0 Re(z)

0 Re(z)

−2

5   MC  The graph representing the region described by {z: | z + 1 | ≥ 3} is: B Im(z) A C Im(z) Im(z) 3 −1 + 0i 0 1 Re(z) −4

3

1 + 0i −2 0 1 4 Re(z) −3

−3

Region required

Region required

D

Im(z)

3

−1 + 0i Re(z) –4 –1 0 2 –3

Region required

3 −1 + 0i Re(z) –4 –1 0 2

E

Im(z)

2

Re(z)

−3 −1 0 0–i

3

−4

Region required

6   WE12  Represent the following regions on the complex plane.

π  a  z: 0 < Arg ( z ) ≤  6 

178

π  − 2π b  z: < Arg ( z ) ≤  3 2 

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

–3

Region required

  −π c  z: ≤ Arg( z ) < π  3  

2π   d z: 0 ≤ Arg ( z + 1) <  3   5π   π f  z: < Arg ( z − 3 + i) <  6 6 

− −π  3π e  z: ≤ Arg( z − 2i) ≤  4 2  7   WE13a  Sketch the graph of each of the following. a {z: Re(z) > −3} b {z: Re(z) ≤ 2} c {z: Im(z) ≥ 1} d {z: Im(z) < −4} e {z: Re(z + 2) ≤ 1} f {z: Re(z − 4) > −3} g {z: Im(z + 3i) > −2} h {z: Im(z − 5i) ≤ 4} i {z: Re(z + 3 − 2i) ≥ 3} j {z: Im(z − 1 + 4i) < −3}

8   WE13b  Sketch the graph of each of the following. a {z: −1 < Re(z) < 5} b {z: −6 < Im(z) ≤ −3} c {z: 0 < Re(z − 3) < 1} d {z: 2 < Im(z + 2i) < 4} e {z: −2 < Re(z − 1 + 2i) ≤ 3} f {z: −4 ≤ Im(z − 3 + 3i) < 0} 9   WE14a  Illustrate each of the following on an Argand diagram. a {z: | z | ≤ 2} b {z: | z | > 5} c {z: | z − 2 | < 3} d {z: | z + 3i | ≥ 1} e {z: | z − 3 + 2i | ≤ 4} f {z: | z + 5 − i | > 3} 10   WE14b  Illustrate each of the following on an Argand diagram. a {z: 1 < | z | ≤ 3} b {z: 3 ≤ | z | < 4} c {z: 2 < | z − 1 | ≤ 5} d {z: 1 < | z − 2i | < 2} e {z: 3 ≤ | z − 2 + 3i | ≤ 4} f {z: 4 ≤ | z + 3 − 4i | < 6} 11   WE15  Sketch the graph of each of the following. 3π   −π < Arg( z ) ≤  ∩ {z: | z | ≤ 2} a  z: 4  2 b {z: Re(z) ≥ 3} ∩ {z: Im(z) < −2}

π  −π ≤ Arg( z ) <  ∩ {z: Re( z ) < 2} c  z: 2  4 2π   π d {z: | z + 3 | > 2} ∩  z: < Arg ( z + 3) ≤  6 3   e {z: Im(z) > 3} ∩ {z: 1 ≤ | z − 2i | < 2} f {z: −4 < Re(z − 3) ≤ −2} ∩ {z: 1 < Im(z − 1) ≤ 3}

4d

Graphs of other simple curves The types of graphs we have looked at up to now can be sketched from the complex relation by recognition. These include rays, vertical lines, horizontal lines, circles and ellipses. However, if the given complex relationship is unfamiliar then it should be converted to the more recognisable Cartesian form and graphed accordingly. We have already seen that: z = x + yi and x = Re(z) y = Im(z) x2 + y2 = | z | We will now consider the translations of these expressions in the following worked example.

Chapter 4  Relations and regions of the complex plane

179

Worked Example 16

Express each of the following expressions in Cartesian form. a   Re(z + 5) b   Im(z − 2 − 3i) c   | z − 4 + 2i | d   Repeat parts a , b and c using a calculator. Think a b

c

d

Write

1

Replace z with x + yi.

2

Write down the real component only.

1

Replace z with x + yi.

2

Write down the imaginary component only.

1

Replace z with x + yi.

2

Collect the real and imaginary components separately.

= | (x − 4) + (y + 2)i |

3

Take the absolute value of the expression.

= ( x − 4)2 + ( y + 2)2

1

First, the settings need to be changed. Tap: d •  O •  Basic Format Under Advanced check the Complex Format and Variable is Real. On the Main screen, complete the entry line as: x + yi W z Then press E. To answer part a, tap: •  Action •  Complex •  re Complete the entry lines as: re(z + 5) im(z - 2 - 3i) abs(z - 4 + 2i) Press E after each entry.

2

Write the answers.

a

Re(z + 5) = Re(x + yi + 5) =x+5

b Im(z − 2 − 3i) = Im(x + yi − 2 − 3i)

=y−3 c | z − 4 + 2i | = | x + yi − 4 + 2i |

For a , Re(z + 5) = x + 5. For b , Im(z - 2 - 3i) = y - 3. For c , | z − 4 + 2i | = x 2 − 8 x + 4 y + 20 .

To convert a complex equation to a Cartesian equation, use: 1. Re(z + a + bi) = x + a 2. Im(z + a + bi) = y + b 3. | z + a + bi | = ( x + a)2 + ( y + b)2 (where a, b ∈ R), then 4. manipulate the equation until the type of curve is recognised. Note: A CAS calculator can be used to obtain the graph in Cartesian form if it is not recognised after manipulating the equation.

180

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

Further straight line graphs WoRked example 17 a   Find the Cartesian equation for 5Re(z) − 4Im(z) = 20. b Sketch the graph on the complex plane. Think

WRiTe a 5Re(z) – 4Im(z) = 20

a Replace Re(z) by x and Im(z) by y to express in

5x − 4y = 20

Cartesian form. b 1 This is a straight line with an x-intercept of 4 (real axis) and a y-intercept of −5 (imaginary axis). 2

b

Im(z) 0

4 Re(z)

Sketch the straight line on the complex plane. –5

WoRked example 18

eBook plus

a   Find the Cartesian equation for | z + 2 − i | = | z − 1 |. b Sketch the graph on the complex plane.

Tutorial

int-0377

Think a

1

WRiTe

Replace | z + 2 − i | by ( x + 2)2 + ( y − 1)2

| z + 2 − i| = |z − 1 |

a

( x + 2)2 + ( y − 1)2 = ( x − 1)2 + y 2

and | z − 1 | by ( x − 1)2 + y 2 to express in Cartesian form.

b

Worked example 18

(x + 2)2 + (y − 1)2 = (x − 1)2 + y2

2

Square both sides to remove the square root signs.

3

Expand the brackets.

4

Subtract x2, y2 and 1 from both sides.

4x + 4 − 2y = −2x

5

Collect like .

6x + 4 − 2y = 0

6

Divide the equation by a common factor of 2.

7

Express the equation in a recognisable form.

1

The equation is a straight line with a y-intercept of 2 and gradient 3. Sketch the straight line on the complex plane.

2

x2 + 4x + 4 + y2 − 2y + 1 = x2 − 2x + 1 + y2

3x + 2 − y = 0 y = 3x + 2 b Im(z) 2 −2– 3

0 Re(z)

Reciprocal graphs WoRked example 19

If Im(z) Re(z) = 4, find its Cartesian equation and sketch the graph on the complex plane. Think 1

WRiTe

Replace Im(z) with y and Re(z) with x to express in Cartesian form.

Chapter 4

Im(z) Re(z) = 4 yx = 4

Relations and regions of the complex plane

181

2

Divide both sides by x.

3

The equation represents a reciprocal function with asymptotes through x = 0 and y = 0 and ing through (1, 4) and (−1, −4) on a Cartesian plane.

4

Sketch the graph on the complex plane.

y=

Im(z)

4 x

1 + 4i

0

Re(z)

−1 − 4i

Graphs of parabolas Worked Example 20

For Im(z) = [Re(z − 1)]2 find the Cartesian equation and sketch its graph on the complex plane. Think 1

Replace Im(z) with y and Re(z − 1) with x − 1 to express in Cartesian form.

2

The equation represents a parabola with a minimum turning point (1, 0) and y-intercept (0, 1) on a Cartesian plane.

3

Sketch the graph on the complex plane.

Write

Im(z) = [Re(z -1)]2 y = [Re(x + yi - 1)]2 y = (x − 1)2

Im(z) 1 –1

–1

0 1 Re(z)

Worked Example 21

Show that the locus described by |z| = Im(z) + 1 is a parabola. Write the parabola’s Cartesian equation and state its turning point and type. Think

Let z = x + yi, and express the equation in of x and y.

Let z = x + yi

2

Square both sides and expand the brackets.

x2 + y2 = (y + 1)2 x2 + y2 = y2 + 2y + 1

3

Simplify.

x2 = 2y + 1

4

Make y the subject.

y=

5

State the turning point and its type.

There is a minimum turning point at −1 (0, 2 ) .

1

182

Write

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

x2 + y2 = y + 1

1 2 1 x − 2 2

Regions involving simple curves WoRked example 22

eBook plus

Sketch the following on the complex plane. a {z: Im(z) − 2 < [Re(z + 3)]2} b {z: Im(z) Re(z) ≥ −1} ∩ {z: Im(z) < 2} Think a

b

1

int-0378 Worked example 22

WRiTe

Replace Im(z) with y and Re(z + 3) with x + 3 to express in Cartesian form.

a Im(z) − 2 < [Re(z + 3)]2

y − 2 < (x + 3)2

y < (x + 3)2 + 2

2

Add 2 to both sides.

3

The boundary curve of the inequation represents a parabola with a minimum turning point (−3, 2) on a Cartesian plane.

4

Sketch the parabola with a broken line on the complex plane and shade the region above the parabola leaving the unshaded area as the region required.

1

Tutorial

Replace Im(z) with y and Re(z) with x to express the first inequation in Cartesian form.

Im(z) 2 Region required

0 Re(z)

−3

b Im(z) Re(z) ≥ −1

yx ≥ −1 y≥

2

Divide both sides by x.

3

The boundary curve of the inequation represents a negative reciprocal function with asymptotes x = 0 and y = 0 and ing through the points (1, −1) and (−1, 1) on a Cartesian plane.

4

Sketch the reciprocal function with a solid line on the complex plane and shade out the region below each branch of the curve.

−

1 x

Im(z)

−1 + i

Re(z) 0 1−i

5

6

7

The second inequation represents a half-plane below the horizontal line ing through 0 + 2i.

Im(z) −1 + i 2

Sketch this straight line with a broken line on the same complex plane and shade the region above the line. Show the unshaded region as the region required.

Re(z) 0 1−i

Im(z) −1 + i

2 Re(z) 0

Region required

Chapter 4

1−i

Relations and regions of the complex plane

183



To convert a complex equation to a Cartesian equation, use: 1. Re(z + a + bi) = x + a 2. Im(z + a + bi) = y + b 3. | z + a + bi | = ( x + a)2 + ( y + b)2    where a, b ∈ R then manipulate the equation until the type of curve is recognised.

Exercise

4d

Graphs of other simple curves A CAS or graphics calculator can be used where necessary to assist with the sketch graphs in the following exercise.

1   WE 16  Express each of the following expressions in Cartesian form. a Re(z − 8) e | z + 3 | i Re(z − 7 + 3i)

b Im(z + 2i) f | z − 5i | j | 5 − 4i − z |

c Re(z + 5 − 4i) g | z + 4 + 6i |

d Im(z − 2 − 3i) h | z − 1 + 5i |

2   WE 17  For each of the following equations: a d

i express the equation in Cartesian form ii sketch its graph on the complex plane. Re(z) + Im(z) = 4 b 2Re(z) − Im(z) = 6 1 Im(z) = 4Re(z) + 1 e Im(z) = 2 Re(z) − 3

c 3Re(z) + 2Im(z) = 6 f 3Im(z) − 4Re(z) = 12

3   WE 18  For each of the following equations: i express the equation in Cartesian form ii sketch its graph on the complex plane. a | z + 2 | = | z + i | b | z − 3 | = | z + 3i | c | z − 1 + 2i| = | z + 3 + i | d | z − 2 − 3i | = | z + 1 − 4i | e | z + 3 − 2i| = | z + 3 + 3i | f | z + 4 − 3i | = | z + 2 − 3i | 4   MC  The rule which describes the relationship shown by the graph at right is: A 3Im(z) − 2Re(z) = 6 B 2Im(z) − 3Re(z) = 6 C 3Re(z) − 2Im(z) = 6 D Im(z) = 32 Re(z) E 2Re(z) − 3Im(z) = 6 5   MC  The Cartesian equation which represents | z + 2 − i | = | z + i | is: A y = x + 1 B y = x − 1 C x + y = 1 D y = 1 − x E y2 = x2 + 1 6   MC    a  The Cartesian equation which describes the graph shown at right is:  −1  A y =    x

 −1  B y =  2   x 

 −2  C y =    x

 2 D y =  2  x 

 −2  E y =  2   x 

184

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

Im(z) 3

0 Re(z)

−2

Im(z) Re(z) 0 −1 − 2i

1 − 2i

b The relationship described in complex form is: B Im(z) [Re(z)]2 = 2 A Im(z) Re(z) = −2 D Im(z) = −2[Re(z)]2 E Im(z) = [−2Re(z)]2

C Im(z) [R(z)]2 = −2

7

We19 Find the Cartesian equation and sketch the graph on the complex plane for each of the following. a Im(z) Re(z) = 2 b Im(z) Re(z) = −2 c Im(z) Re(z − 1) = 3 d Im(z) (Re z)2 = 1 2 − e Im(z) (Re z) = 4 f Im(z) [Re(z + 1)]2 = 2

8

We20 Find the Cartesian equation and sketch the graph on the complex plane for each of the following. a Im(z) = [Re(z)]2 b Im(z) = [Re(z)]2 + 2 − 2 c Im(z) = 2[Re(z)] − 1 d Im(z) = [Re(z − 4)]2 − 2 e Im(z + 1) = [Re(z + 2)] f Im(z − 3i) = 2[Re(z + 1)]2 + 2 2 g Im(z + 2i) = 1 − [Re(z − 2)]

9 Sketch the following complex relations on a complex plane. a Im(z) = [Re(z)]3 b Im(z − 2i) = [Re(z + 1)]3 c zz =4 d zz =3 | z +3| |z+i| e = 2 f |z− 3i| = 2 | z −1| 2 10 We21 Show that the locus described by each of the following is a parabola. i Write the Cartesian equation. ii State the turning point and its type. a | z | = Im(z) − 1 b | z | = Im(z) + 2 c | z | = Im(z) − 3 11 We22 Sketch the following on the complex plane. a {z: Im(z) ≥ [Re(z)]2} b {z: Im(z) Re(z) ≤ −4} c {z: | z + 3| ≤ | z − 3i |} d {z: 5Im(z) − 4Re(z) ≤ 10} e {z: Im(z) > [Re(z − 2)]2} ∩ {z: Re(z) ≤ 2} f {z: Im(z) [Re(z)]2 ≥ 1} ∩ {z: Im(z) > Re(z)} g {z: −Im(z) Re(z + 1) > 1} ∩ {z: Im(z − 2i) < 1 + [Re(z − 1)]2}

eBook plus Digital doc

WorkSHEET 4.2

Chapter 4

Relations and regions of the complex plane

185

Summary Rays and lines

• The graph of {z: Arg(z) = θ} is a ray starting at the origin and making an angle of θ with the positive real axis. • The graph of {z: Arg(z + a + bi) = θ} is the same as {z: Arg(z) = θ} translated −a units in the real direction and −b units in the imaginary direction. • The graph of {z: Re(z + a + bi) = c} is a vertical line ing through the point (c − a) + 0i. • The graph of {z: Im(z + a + bi) = c} is a horizontal line ing through the point 0 + (c − b)i. Circles and ellipses

• The relation {z: |z + a + bi| = r} defines a circle of radius r and centre −a − bi. [Re( z ) − 4] 2 [Im( z ) − k ]2 + = 1 is an ellipse with centre h + ki and semi-major axis a2 b2 length of a and semi-minor axis length of b. • {z: | z − A |+ | z − B | = C} describes an ellipse where A and B are two points in the complex plane and C is a real constant. • The complex equation

Combination graphs and regions

• More than one graph or region can be represented on the complex plane. • For regions: 1. The boundaries are represented by: (a)  solid lines where ≤ or ≥ apply (b)  broken lines where < or > apply. 2. Shade the regions not required. • Wherever practicable, check that the region required is correct by selecting a point in the region and make sure that it satisfies the inequation(s). Graphs of other simple cur ves

• To convert a complex equation to a Cartesian equation use: 1. Re(z + a + bi) = x + a 2. Im(z + a + bi) = y + b (where a, b ∈ R), then 3. | z + a + bi | = ( x + a)2 + ( y + b)2 4. manipulate the equation until the type of curve is recognised.

186

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

chapter review 10 A graph is defined as {z: | z | = Im(z) + 2}. a Write the equation in Cartesian form. b Describe the locus of the graph. c State the turning point and its type. d Sketch the graph {z: | z |< Im(z) + 2}. π 11 a   i Let z1 = cis   . Plot and label carefully the  4

Short answer 

1 Sketch the graph of {z: Arg(z − 5 − 3i) = an Argand diagram.

− 2π

5

} on

2 Sketch the graph of {z: Im(z + 20 + 3i) = −1} on the complex plane. 3 Illustrate {z: | z − 3 + 3i | = 1} on the complex plane.

 points -z1, z1 and z1 on the Argand diagram below.

4 Represent {z: 4[Re(z + 1)]2 + 3[Im(z − 2i)]2 = 12} on an Argand diagram. 5 Write down the complex relation that describes the graph shown.

Im(z)

Im(z) 5

−

2 1

3

π 2

−3 −2 1

−3 −2 −1

−1

0

1

2

3Re(z)

−1 0 1 2 3 Re(z)

−2

6 Show {z: Im(z + 5i) < 4} ∩ {z: 1 < | z − 2 + 2i | ≤ 3} on an Argand diagram.



7 Express the complex relation | z − 4 | = | z + 2 − 3i | in Cartesian form. 8 Show the graph of {z: Im(z + i) Re(z − 3) > 2} ∩ {z: Im z ≥ −[Re(z + 1 + 3i)]2} on the complex plane.

b Use a double angle formula to show that the

9 On the complex plane: a sketch the graph {z: Im(z) Re(z) = −1} b write the Cartesian equation of the graph. c From part a, describe the series of translations required to sketch the graph below.

2+ 2 π exact value of cos   = . Explain  8 2 why any values are rejected. c Hence show that the exact value of π sin   =  8

Im(z) 3 2

−3 −2 −1

ii Write down the complex equation of the straight line which es through the points z1 and -z1, in of z1.

2− 2 . 2

Exam tip   In this case, ‘show that’ will require the correct steps in manipulating surds. Also, the ‘hence’ instruction implies that the answer for

0 1 2 3 4 5 6 Re(z)

π cos   is to be used.  8

©VCAA Assessment report 2006

[©VCAA 2006]

d Write the equation of the graph. e Write the Cartesian equation of graph. f Sketch the region for {z: Im(z + 2) Re(z + 2 ) ≥ −1}.

7

 2+ 2 2− 2  d Evaluate  + i ,   2 2 giving your answer in polar form.

Chapter 4  Relations and regions of the complex plane

187

e For what values of n is

Multiple choice n

 2+ 2 2− 2  + i a real number?    2 2 f Plot the roots of z8 = 1 on the Argand diagram below.

1 The graph that correctly represents −π   z: Arg ( z − 4i) = 5 

A Im(z)

Exam tip   Make sure that you do not use rays

B Im(z)

π – 5

4

from the origin to represent complex numbers. Just plot the points.

  is: 

0 −4

©VCAA Assessment report 2006

[©VCAA 2006]

Im(z)

0

−3 −2

−1

0

1

2

3Re(z)

0

4

– −π 5

E Im(z) 4

− –π 5

Re(z)

−2

0 [©VCAA 2006]

12 The cubic equation + + bz + c, where a, b and c are real numbers, has three roots; two of them are 5 and −1 − 3i. a State the third root. b Find the values of a, b and c. c Plot the three roots on an Argand diagram. Label the real root R and the complex roots P and Q. d State the equation of the line ing through the complex points P and Q. e If point M is the midpoint of line PQ, what is the equation of the circle with centre M whose circumference es through P and Q? z3

Re(z)

Re(z) − –π 5

−4

−1

az2

13 For the equation z2 + 10z + 169 = 0, where z ∈ C, find: a the roots of the equation b the sum of the roots c the product of the roots. d Using your results from parts b and c, find the quadratic equation which has roots 3 + 2i and 3 − 2i. 14 Factorise P(z) = −9z3 + 9iz2 − 4z + 4i, where z ∈ C, and hence state the solutions to P(z) = 0. 15 For the complex numbers u = 3 + i and v = 3 − 3i, a write u and v in polar form b find u3v2 in polar and Cartesian form c solve z4 = u3v2, where z ∈ C. 188

D Im(z) 0

1

– −π 5

Re(z)

C Im(z)

2

Re(z)

2 The straight line shown below can be described by: Im(z)

−7

0

Re(z)

B {z: Re(z) = −7} D {z: Im(z) = 7}

A {z: Re(z) = 7} C {z: Im(z) = −7} E {z: Re(z − 7) = 0}

3 The ray shown below can be described by: Im(z) π 3— 4

−3 + 4i 0 Re(z)

3π   A  z: Arg( z − 3 + 4i ) =  4   3π   B  z: Arg( z − 3 ) =  4   3π   C  z: Arg( z + 3 − 4i ) =  4  

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

E Im(z)

3π   D  z: Re( z + 3 − 4i) =  4  

3 2 1

− 3π    E  z: Arg( z + 3 − 4i) =  4   4 The Argand diagram below that correctly represents {z: | z − 2i | = 4} is:

A

C

0 + 2i 0

Im(z) 2

4 Re(z)

−4

D

0

−2

2 Re(z)

Im(z)

−2

π  C { z: | z + 2 | = 4} ∪  z: Arg ( z − 1) =  6  π  D { z: | z + 2 | = 2} ∪  z: Arg ( z + 1) =  6 

2 + 0i 0 2 4 Re(z)

π  E { z: | z − 2 | = 4} ∪  z: Arg ( z − i) =  6  8 The region shown can be described by:

5 The circle below is described by the relation:

Im(z)

Im(z)

Region required

1

−5

–2

0 1 Re(z)

π  B { z: | z + 2 | = 2} ∪  z: Arg ( z − 1) =  6 

2 0 + 2i

−6

–4 –2

π – 6

π  A { z: | z − 2 | = 2} ∪ z: Arg ( z ) =  6 

Im(z) 4

−4 −2 0 4 Re(z) 0 − 2i

2

Re(z)

7 The graph shown below can be described by:

0

−4

4 Re(z)

Im(z) 2

E

1– 2

4

2 −2

0

B Im(z)

Im(z) 6

−4

– 1– 2

−3 − i

Re(z) −1 0

3π –– 7

0

5π − –– 8

Re(z)

−3

A {z: | z − 3 − i | = 2} C {z: | z + 3 − i | = 4} E {z: | z + 3 + i | = 4}

B {z: | z − 3 − i | = 4} D {z: | z + 3 + i | = 2}

 − 5π 3π   B  z: < Arg ( z ) < 8 7  

6 The graph below that correctly represents the relation {z: 4[Re(z)]2 + [Im(z + 2i)]2 = 1} is: B Im(z) A Im(z) 0 − 1–2

−1 −2

1– 2

Re(z)

Re(z)

−2

−3

C Im(z) 0 − 1–2

−2 −4

1– 2

D

–1 –2 –3

0

2

Im(z)

Re(z)

1

− 5– 2

 − 5π 3π   ≤ z: Arg ( z ) ≤ A  z: 8 7  

Re(z) −2− 3– −1 0 2 −1

−  3π 5π   C  z: ≤ Arg ( z ) < 7 8   − 5π   3π     ∪  z: Arg( z ) ≥ D  z: Arg ( z ) < 7  8   

 − 5π 3π   < z: Arg( z ) ≤ E  z: 8 7   9 The graph that represents the region described by 3π   {z: | z + i | > 3} ∩  z: 0 ≤ Arg ( z − 3i) <  is: 4 

Chapter 4  Relations and regions of the complex plane

189

A

Im(z) 2 −3

B

π 3–– 4

Re(z) 3

−1 0 −4 Im(z)

D

π 3— 4

−3

3

−4

Region required

Im(z) 3 2 −3

−1 0 −2 3––π −3 4 −4

Im(z)

D

Re(z) 0 1−i

Re(z)

1 + 3i

2 0 Re(z)

E Im(z) 0 1−i

3

Region required

π 3— 4

3+i 2 Re(z)

13 The region represented on the Argand diagram below, where a is a real constant, could be defined by: Im(z)

Re(z)

−1 0

Im(z) −1 + 3i

−2 −1 − 3i

3

Re(z)

Re(z) −3 −1 0

C

Im(z) 2

2

E

π 3–– 4

Region required

Region required

C

Im(z) 4 3 2 1 –3 −1 0 −2

Region required

a + 2i

3

0

−4

Re(z)

Region required

10 The complex equation | z − 3 + 2i | = | z + 3 | can be expressed in Cartesian form as: A y = −1 B y = x2 + 1 C y = 3x − 1 D y = 3 − x E y = 1 − 3x 11 The graph shown can be described by:

0

4Re(z)

−4

A y = −2x B y = 2x − 4 D 2x + y + 4 = 0 E y = 4 − 2x 12 The graph that represents {z: Im(z − 2i) Re(z) = 1} is: A

B

Im(z) 2 −1 + i

C 2y + x = −4

Im(z)

1 + 3i

0 Re(z)

1+i −1 − i

B | z - (a + 2i) | ≤ 1 D | z - (a + 2i) | ≤ 2



[©VCAA 2006]

14 Which one of the following relations does not have a graph that is a straight line ing through the origin? A z + z = 0 B 3 Re(z) = Im(z) C z = zi D Re(z) - 2 Im(z) = 0 E Re(z) + Im(z) = 1 [©VCAA 2006]

Im(z) −2

A | z - (a + 2i) | ≥ 1 C | z - (-a + 2i) | ≤ 1 E | z + a - 2i) | ≥ 1

0 Re(z)

15 The relation ( z + 2)( z + 2) = 4, when graphed on an Argand diagram, would be: A a circle of radius 4 with centre at (-2, 0) B a circle of radius 2 with centre at (2, 0) C a circle of radius 4 with centre (2, 0) D a circle of radius 2 with centre (-2, 0) E a circle of radius 2 with centre at (0, -2)

[©VCAA 2008]

Extended response

1 Using z = x + yi sketch each of the following on separate complex planes. a | z − 1 − i | = 1 b | iz − 1 − i | = 1 d | i3z − 1 − i | = 1 e | i4z − 1 − i | = 1 Comment on the result.

190

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

c | i2z − 1 − i | = 1

2 a   Find the Cartesian equation of each of the following. i | z − 3 | = | 2z | ii | z + 16 − 16i | = | 3z | b Without graphing show that the two graphs do not intersect. c Find the exact area of the region defined by {z: | z − 3 | ≤ | 2z |} ∩ {z: | z + 16 − 16i | ≥ | 3z |}. 3 Let z1 = 1 + 2i and z2 = -2 - i. a Show z1 and z2 on an Argand diagram. b Draw a straight line ing through z1 and z2, and find its Cartesian equation. c Express the complex relation | z - 1 - 2i | = | z + 2 + i | in Cartesian form. d Add the graph of the complex relation | z - 1 - 2i | = | z + 2 + i | to the Argand diagram in part a. e Prove that the locus of the points described by the complex relation | z - 1 - 2i | = | z + 2 + i | is a perpendicular bisector of the line segment ing z1 and z2. f Generalise this result for any two complex numbers z1 and z2. 5 4 a   Find the Cartesian equation of the complex relation desribed by | z − 4 | + | z + 4 | = . 2 b State the domain for the relation in a. c Identify the shape of the locus of the points, described by the equation in a. d Sketch the graph of the above relation on an Argand diagram. 5 a   If z = x + yi, find the Cartesian equation defined by the complex relation zz = 16. b Identify the curve described in a, and sketch it on an Argand diagram. c Find the Cartesian equation defined by the relation zz = a , where a ∈ R+. d Identify the curve described in c, and hence complete the following statement: ‘If z = x + yi, the graph of zz = a , where a ∈ R+ is a .............. .’ e Find the Cartesian equation defined by the relation [ z − (2 + 3i)][ z − (2 − 3i)] = 4. f Identify the curve described in e, and sketch it on an Argand diagram. g If z = x + yi and w = u + vi, where x, y, u, v, ∈, R, find the Cartesian equation defined by the complex relation ( z − w)( z − w) = a , where a ∈ R+. h Identify the curve in g, and hence complete the following statement: ‘If z = x + yi and w = u + vi, where x, y, u, v, the graph of ( z − w)( z − w) = a , where a ∈ R+ is a ........ .’ 6 a    that

3 2

1

+ 2 i is one root of the equation z3 = i.

b Plot the three roots of i on the Argand diagram below. Im(z) 2 1 −3 −2

−1

0

1

2

3 Re(z)

−1 −2

c Find the points of intersection of the curves given by | z - i | = 1 and Re( z ) =

−

1 3

Im( z ).

Chapter 4  Relations and regions of the complex plane

191

d Sketch the curves given by the relations | z − i | = 1 and Re( z ) =

−

1 3

Im( z ) on the Argand diagram below.

Im(z) 2 1 −3 −2

−1

0

1

2

3 Re(z)

−1 −2

{

e On the Argand diagram above shade the region given by {z: | z − i | ≤ 1} ∩ z: ≤ Arg( z ) ≤ f Find the area of the shaded region in part e, correct to two decimal places.

2π 3

}. [©VCAA 2008]

eBook plus Digital doc

Test Yourself Chapter 4

192

maths Quest 12 Specialist mathematics for the Casio Classpad

eBook plus

aCTiviTieS

Chapter opener Digital doc

• 10 Quick Questions: Warm up with ten quick questions on relations and regions of the complex plane. (page 162) 4B

Circles and elllipses

Tutorial

• We7 int-0374: Watch a tutorial on how to express a complex relation as a Cartesian equation and sketch the resultant graph. (page 169) • We10 int-0375: Watch a tutorial on how to describe and sketch a complex relation on a complex plane. (page 171) Digital doc

• WorkSHEET 4.1: Represent relations and regions in the complex plane. (page 174) 4C

Combination graphs and regions

Interactivity

• Combination graphs and regions int-0345: Consolidate your understanding of relations and regions of the complex plane. (page 174) Tutorial

• We12 int-0376: Watch how to represent a region defined by a complex inequation on a complex plane. (page 175) 4D

Digital docs

• WorkSHEET 4.2: Sketch graphs in the complex plane. (page 185) Chapter review Digital doc

• Test Yourself: Take the end-of-chapter test to test your progress. (page 192) To access eBookPLUS activities, log on to www.jalus.com.au

Graphs of other simple curves

Tutorial

• We18 int-0377: Watch a tutorial on sketching regions on a complex plane. (page 181) • We22 int-0378: Watch a tutorial on sketching regions on a complex plane. (page 183)

Chapter 4

Relations and regions of the complex plane

193

EXAM PRACTICE 1 Short answer



25 minutes

1 a Express 5 − 3i in the form a + bi. 2 marks 2−i b Write down the polar form for -6 + 2 3i. (Note root 3.) 2 marks 2 2 a + bi is a solution of 5x - 3x + 4 = 0. Write down the second solution in of a and b. 1 mark 3 a Show that z = 1 - 2i is a solution of z3 + (1 + 2i) z2 - (1 - 4i)z - 1 + 2i = 0. b Find the remaining solutions.  3π  1 4 a Given that cot (x) = 2 and x ∈  π ,  ,  2 determine the value of sin (x). b Hence find an exact value for cos (2x).

4 marks

5 The graph below is of y = a sec b (x + c) + d. Evaluate a, b, c and d.

Multiple choice

1 An ellipse has the parametric form x = 1 + 3 cos (t) and y = -2 + sin (t). The coordinates of the centre of the ellipse and the horizontal semi-axis length are: c (-2, 1), 1 a (1, 2), 3 B (1, -2), 1 -2), 3 E (1, 2), 9 D (1, 2 A hyperbola has an asymptote of y = -2x + 3. The asymptotes intersect at x = 2. The Cartesian equation of the hyperbola is: a

( x − 2)2 ( y + 1)2 ( x − 2)2 ( y + 1)2 − = 1 B  − =1 2 1 4 16

c

( x − 2)2 ( y − 1)2 ( x + 1)2 ( y − 2)2 − = 1 d  − =1 1 2 4 16

e

( x + 2)2 ( y + 1)2 − =1 4 16

c − a  1 − , 1 − b 

−1 −0.6 −0.2 0 0.2 −1 −2

0.6

1

1.8 x

1.4

12 minutes

Each question is worth one mark.

3 The domain of the function f : D → R where f (x) = aarcsin (bx + c) + d is:

y 4 3 2 1 −1.8 −1.4

Chapters 1 TO 4

4 marks

6 The complex number z has been plotted on the Argand diagram below. Plot and label points corresponding to: a z2 b z c z

c b 

c  − 1 − c, − 1 + c 

 −π a  πa + d, + d . B  2 2    −π  π + d , + d . d  2 2  

 −c − 1 −c + 1 , e   b   b 4 The following diagram illustrates all the roots of a polynomial equation. Im(z)

Im(z) 4 3 2 1 −4 −3 −2 −1 0 −1 −2 −3 −4



194

Re(z)

1 2 3 4 Re(z)

Which of the following could be the equation if {a, b} ∈ R+? a z3 = a B  z3 + a + bi = 0 3 c z + z - a = 0 d  z3 + (a + bi)z2 + z + 1 = 0 e (z2 - a) (z2 + ax + b) 3 marks

Maths Quest 12 Specialist Mathematics for the Casio ClassPad

5 Which of the following expressions correspond to a hyperbola in an Argand diagram? A zz = 1 B ( z − 3i)( z + 3i) = 1 C |z − 3| − |z + 2| = 1 D | z − 3 | − | z + 2i | = 0 E

1 =1 z

1 is equal to: 1− z −1 + i −1 − i 5 − 3i A B C 6 6 34 5 + 3i 4 + 3i D E 34 25  − 2π  3 7 If z = 8 cis   then z equals:  3 6 If z = 4 − 3i then

 2π  A −2 cis    9

B

 − 2π  8 cis  3  9 

 − 2π  D 2 cis  3  3  

 − 2π  C 2 cis   9  E 2

A B 8 The partial fraction can be written + x − c 2 x +c as: a A 2 x 2 − c2 ax + b B ( x − c)(2 x + c) ax 2 + bx C ( x − c)(2 x + c) D

a ( x − c)(2 x + c)

ax + b 2 x 2 − c2 where A ≠ a, B ≠ b and a, b, A, B and c ∈ R. E

exTended ReSponSe

18 minutes

1 a cis (θ ) = cos (θ ) + i sin (θ ) Show that cis (θ ) cis (ϕ) = cos (θ ) cos (ϕ) − sin (θ ) sin (ϕ) + i [cos (θ ) sin (ϕ) + sin (θ ) cos (ϕ)]. b Hence obtain an expression for cos (θ + ϕ). c Show that cos (2θ ) = 2 cos2 (θ ) − 1. d Hence obtain an exact expressions for cos (a) if cos (2a) = k. e Obtain an expression for cos (4θ ) in of cos (θ ) only. 2 a z1 = −1 − i and z2 = −z1 are plotted below. Plot and label z1 on the diagram. b Express z1 in polar form. − π c Show that z = 2 cis   satisfies the complex  4 relationship. | z − z1 | + | z − z2 | = 4. d Plot an additional point on the diagram above which satisfies the complex relationship. | z − z1 | + | z − z2 | = 4. e The Cartesian equivalent relationship is 3x2 + 3y2 − 2xy = k. Evaluate k. f Sketch the region defined by {| z − z1 |<| z − z2 | ∩ | z − z1 | + | z − z2 | ≤ 4}.

5 marks

Im(z) 2 z2

1

−2

−1

0

1

2 Re(z)

−1 z1

−2

1 + 1 + 1 + 2 + 1 = 7 marks

eBook plus Digital doc

Exam practice 1 Solutions

exam practice 1

195