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NELSON SENIOR MATHS SPECIALIST 11 FULLY WORKED SOLUTIONS

CHAPTER 10: Complex numbers Exercise 10.01: Imaginary numbers

Concepts and techniques

1 a 225 25 5− = =i i

b 236 36 6− = =i i

c 220 20 2 5− = =i i

d 28 8 2 2

9 9 3− = =

i i

2 a 2 1x = −

2 2== ±

x ix i

b 2 9= −x

2 29

3== ±

x ix i

c 2 49x = −

2 249

7== ±

x ix i

d 2 14

x = −

22

4

2

=

= ±

ix

ix

© Cengage Learning Australia 2014 ISBN 9780170251501 1

3 a 6i = (–1)3 = –1

b 11i = 10 5( 1)= − = −i i i i

c ( )1428 2 14( 1) 1= = − =i i

d 97 96 48( 1)= = − =i i i i i

4 a ( )i i× − = –i2 = –(–1) = 1

b ( )i i− × − = i2 = –1

c ( )7i− = –i(–i)6 = –i (i6) = –i (–1)3 = –i

d 5 9i i× = 45 i2 = –45

5 a ( )( )1 1i i+ − = 1– i2 = 1 + 1 = 2

b ( ) 23 2 5 6 15 6 15+ = + = − +i i i i i

6 2 3 4 51 1 1 1 1+ − + − + = + + − − + = +i i i i i i i i i

7 ( )( )2 2 216 16 4 4+ = − = − +x x i x i x i


Reasoning and communication

8 a ( ) ( )2 3 4 100... 1 1 1 1 ...25 timesi i i i i i i i i+ + + + + = − − + + − − + +

= 25(0)

= 0

b 253

0 1 2 253

0.....

=

=

= + + +∑n

n

ni i i i i

( ) ( )

1 2 2531 .....1 1 1 1 .... 1

63(0) 11

i i ii i i i i

ii

= + + + +

= + − − + + − − + + +

= + += +

9 a 4

3 21= = = = −

i i i i ii i

b 4

3 3

1= =

i ii i

c 4

4 4

1 1=ii i

d 5 5 4 1 1 4 3 2− − − −= × = = × = = = −i i i i i i i i i i

e 10 10 4 4 4 2 1− −= × × × = = −i i i i i i

10 a ( )( )2 2 2 24 1( 4) 4 2 2x x x i x i x i+ = − − = − = − +

b ( )( )2 2 2 281 1( 81) 81 9 9+ = − − = − = − +x x x i x i x i


Exercise 10.02: Complex numbers


1 a 5 2i+ Complex

b 5i Purely imaginary

c 22 2i = − Real

d 4 32− Real

e 8− Purely imaginary

f 2 4i i+ = –1 + (–1)2 = 0 Real

g 2i − Complex

h 3 2 3 2−= −

i ii

Real

2 a 2 4z i= − − ( )Re z = –2 ( )Im z = –4

b 7 34

iz += ( )Re z = 7

4 ( )Im z = 3

4

c 2 6z i= − ( )Re z = – 6 ( )Im z = 2

d 2 2 2 2

2 2 2 2 2 2

−= −

+ + +x iy x iyx y x y x y

so ( )Re z = 2

2 2+x

x y ( )Im z =

2

2 2−+y

x y

3 a 5 4w i= − b 2 7w i= +

c 1 32 2

iw = − +

d 2 2 5 2w i= − −

e 2 2

y xiwx y+

=+


4 a 2 49 0x + =

2

2 2

49497

xx ix i

= −

== ±

b 2 1 0x + =

2

2 2

1xx ix i

= −

== ±

c 2 25 0x + =

2

2 2

25255

xx ix i

= −

== ±

d 2 4 5 0x x+ + = 2

2 416 20 42

b b aci xa

− ± −∆ = − = =

24 42

4 22

2 or 2

ix

i

x i x i

− ±=

− ±=

= − + = − −

e 2 2 0z z− + = 2

2 41 8 72

b b aci za

− ± −∆ = − = =

21 72

1 72

7 71 1or2 2 2 2

iz

i

i iz z

±=

±=

= + = −


f 2 2 2 0w w− + = 2

2 44 8 42

b b aci wa

− ± −∆ = − = =

22 42

2 22

1 or 1

iw

i

w i w i

±=

±=

= + = −

g 2 2 4 0x x− + = 2

2 44 16 122

b b aci xa

− ± −∆ = − = =

22 122

2 2 32

1 3 or 1 3

ix

i

x i x i

±=

±=

= + = −

h 2 10 6z z+ =

i.e. 2 6 10 0z z− + = 2

2 436 40 42

b b aci za

− ± −∆= − = =

26 42

6 22

3 or 3

iz

i

z i z i

±=

±=

= + = −

i 2 4 8w w= −

i.e. 2 4 8 0w w− + = 2

2 416 32 162

b b aci wa

− ± −∆= − = =

24 162

4 42

2 2 or 2 2

iw

i

w i w i

±=

±=

= + = −



5 a ( )Re 3z x= + , ( )Im 2z y= −

b ( )Re 3z x y= − , ( )Im 2z x y= +

c 5 4z i x yi= − + −

4 (5 )x y i= − + + −

( )Re 4z x= − + , ( )Im 5z y= −

d ( )4 3 2z x y i x y= − − −

( )4 3 2z x y i x y= − + − +

( )Re 4z x y= − , ( )Im 3 2z x y= − +

e 2 2 2 2

2 2 2 2 2 2 2 2

2 2 21x xyi y x y xyi xyix y x y x y x y+ + +

= + = ++ + + +

( )Re 1z = , ( ) 2 2

2Im xyzx y

=+

6 a 22 3 4 0z z+ + =

2

2

2

9 32 23 23

42

3 234

3 234

3 23 3 23or4 4 4 4

∆ = − = − =

− ± −=

− ±=

− ±=

= − + = − −

i

b b acza

i

iz

i iz z


b 23 6 2z zi+ = i.e. 23 6 2 0z zi+ − =

( ) ( )

2 2

2

2

36 24 12 12

42

6 126

6 2 36

3 3 3 3or

3 3

i i

b b acxa

i i

i ix

i ix x

∆ = + = − =

− ± −=

− ±=

− ±=

− + − −= =

Note that solutions do NOT appear in conjugate pairs because not all of the coefficients of the equation are real. c 2 2 2 2z a b az+ + = i.e. 2 2 22 0z az a b− + + =

2 2 2 2

2

2 2

4 4( ) 4

42

2 42

2 22

a a b b

b b acza

a i b

a ib

a bi

∆ = − + = −

− ± −=

±=

±=

= ±


Exercise 10.03: Operations with Complex numbers


1 a If 1z i= − − , then 1z i= − +

b 32

iz −=

c 3 1z i= − −

d ( ) ( )2 3 2 3− + − = − − −x y x y i x y x y i

e If z = 2 2

2 2

x iyx y++

, then z = 2 2

2 2

−+

x iyx y

2 a ( )( )2 2i i+ − = 4 – i2 = 4 + 1 = 5 which is real

b 25 2 5 2 25 4 25 4 29 17

2 2 4 4 4 4i i i+ − − +× = = = = which is real

c ( ) 2( 5 3) 5 3 5 9 14− − − = − + =i i i which is real

3 2 2 3x x+ + = 2 22 1 2x x i+ + − = (x + 1)2 22− i

= (x + 1 – 2i )( x + 1 + 2i ) 4 a 2 4 5x x+ + = 2 24 4+ + −x x i = (x + 2)2 2−i = (x + 2 – i )(x + 2 + i ) b 2 6 13x x− + = 2 6 9 4− + +x x = (x – 3)2 24− i = (x – 3– 2 i )( x – 3 + 2i) c 2 2 2x x− + = 2 2 1 1− + +x x = (x – 1)2 2−i = (x – 1 – i )( x – 1+ i)


5 a 2 4 0x x+ + =

2

2

2

1 16 15 15

42

1 152

1 152

1 15 1 15or2 2 2 2

i

b b acxa

i

ix

i ix x

∆ = − = − =

− ± −=

− ±=

− ±=

= − + = − −

b 2 3 3 0x x− + =

2

2

2

9 12 3 3

42

3 32

3 32

3 3 3 3or2 2 2 2

i

b b acxa

i

ix

i ix x

∆ = − = − =

− ± −=

±=

±=

= + = −

c x2 + x + 10 = 0

2

2

2

36 40 4 4

42

6 42

6 22

3 or 3

i

b b acxai

ix

x i x i

∆ = − = − =

− ± −=

− ±=

− ±=

= − + = − −


d 22 4 0x x+ + =

2

2

2

1 32 31 31

42

1 314

1 312

31 311 1or2 2 2 2

i

b b acxa

i

ix

i ix x

∆ = − = − =

− ± −=

− ±=

− ±=

= − + = − −

6 a 1 i− ± ( ) ( )1 1 1 0x i x i− − + − − − =

2

2

2 2

2

2 ( 1 )( 1 ) 02 (1 )(1 ) 02 (1 ) 02 2 0

x x i ix x i ix x ix x

+ + − + − − =

+ + − + =

+ + − =

+ + =

b 1 3i± ( ) ( )1 3 1 1 3 0x i x i− + − − =

2

2 2

2

2 (1 3 )(1 3 ) 02 (1 9 ) 02 10 0

x x i ix x ix x

− + + − =

− + − =

− + =

c 3 i± ( ) ( )3 3 0x i x i − + − − =

( )( )2

2 2

2

2 3 3 3 0

2 3 3 0

2 3 4 0

x x i i

x x i

x x

− + + − =

− + − =

− + =



7 a 2 1 0x ix+ + =

2

2

2

4 5

425

2

52

52

∆ = − = −

− ± −=

− ± −=

− ±=

− ±=

i

b b acxa

i

i i

i ix

Note that solutions do NOT appear in conjugate pairs because not all of the coefficients of the equation are real. b 2 2 1 0x ix− − =

( )2 2

2

2 4 4 4 0

42

22

i i

b b acxa

i

x i

∆ = − + = + =

− ± −=

=

=

c 2 2 0ix x i− + =

( ) ( )

( )

( )

2 2

2

1 4 2 1 8 9

42

1 92

1 32

1 321 3

21 3

02

or = 2

i i i

b b acxa

i

ii

i ii

x

ix ,

x i x i

∆ = − − = − =

− ± −=

±=

±=

±= ×

±=

− − ±

=

= −


8 a If the coefficients of a quadratic equation are real, then the roots occur in conjugate pairs. b If the coefficients of a quadratic equation are not all real, then the roots do not occur in conjugate pairs.

9 Given z x yi= +

a Prove ( )2Rez z z+ =

z = x – yi

∴ z z+ = x + x

= 2x

( )2Re z=

b Prove ( )2 Imz z z− =

z = x – yi

∴ z z− =x yi+ x – ( x yi− )

= 2yi

( )2 Im z=


Exercise 10.04: Operations with complex numbers


1 a 2 7 8a bi i+ = −

Equate Real coefficients: Equate Imaginary coefficients: = 7 2 8

47, 4

a bb

a b

= −= −

= = −

b ( ) ( )4 3 2a i b bi i+ − + = −

( ) (4 ) 3 2Equate Real coefficients: Equate Imaginary coefficients:

3 4 26

99, 6

a b b i i

a b bb

aa b

− + − = −

− = − = −=

== =

c 2 3 2 1 5a b ai bi i+ + − = +

Equate Real coefficients: Equate Imaginary coefficients:2 1 3 2 5

1 2 substitute 3 2(1 2 ) 57 7

11

1, 1

a b a bb a a a

aa

ba b

+ = − == − − − =

==

= −= = −

2 a ( ) ( )1 3 2i i− + + = 3 2i−

b ( ) ( )6 5 3 4i i+ − + = 3 i+

c ( )( )3 2 1 9i i− + = 3 + 25i – 18i2 = 21 25i+

d ( ) ( ) 22 3 2 4 7 6 4 4 28 2 24i i i i i i i+ + − = + + − = −


3 Given 2 3z i= + and 5 2w i= − ,

a 4 8 12 5 2 13 10z w i i i+ = + + − = +

b z w− = 3 5i− +

c ( )22 22 3 4 12 9 5 12z i i i i= + = + + = − +

d 2 (2 3 )(5 2 ) 10 11 6 16 11i i iw iz i+ − == + − = +

e 16 11 16 11z w i i= + = −

f ( ) ( ) ( )2 2 2 22 3 5 2 7 49 14 48 14z w i i i i i i+ = + + − = + = + + = +

g 2 2 2(5 2 ) 25 20 4 21 20w i i i i= + = + + = +

h ( ) ( ) 23 3 2 3 5 2 6 9 5 2 4 4z wi i i i i i i i− = + − − = + − + = +

4 a 2

1 1 1 11 1 21

i i ii i i

− − −× = =

+ − − = 1 1

2 2i−

b 2

2

2 2 4 4 3 4 3 42 2 5 5 54

i i i i i ii i i

− − − + −× = = = −

+ − −

c 2

2

3 4 2 3 6 17 12 6 17 6 172 3 2 3 13 13 134 9

i i i i i ii i i

+ + + + − +× = = = − +

− + −

d 2

2

44 1 4 1 44 4 17 17 1716

i ii i i ii i i

++ − +× = = = −

− + −−

e 22 2 1 2 1 2

5 5 5 5 5i i i i i i

i i− − + +

× = = = − −− −

5 Given 1 2u i= − + and 2 3v i= − ,

a 2

1 1 1 1 2 1 2 1 2 1 21 2 1 2 1 2 5 5 51 4

i i i iu i i i i

− − − − − −= = × = = = − −− + − + − − −

b 2

2

1 2 1 2 2 3 2 6 8 8 12 3 2 3 2 3 13 13 134 9

u i i i i i i iv i i i i

− + − + + − + + − += = × = = = − +

− − + −

c

( )2 2 2 2

3 3 3 3 3 4 9 12 9 12 9 123 4 3 4 25 25 251 4 4 9 161 2

i i i ii iu i i ii

− + − + − += = = × = = = − +

− − − +− + −− +


d 2 2

1 1 1 3 2 3 2 3 2 3 23 2 3 2 13 13 132 3 9 4

i i i iiv i ii i i

− − −= = × = = = −

+ −− −

e ( )( )

2 11 2 2 12 2 2 1

iu iv i i i

− −− − += = = −

+ − −

6 a ( ) ( )( ) ( ) ( )2 2 22 3 1 2 1 2 4 4 3 1 4 3 4 3 5 18 4i i i i i i i i− + + − = − + + − = − + = −

b ( )( ) ( )23 5 5 3 3 2i i i+ − − −

( )2 215 25 9 15 9 12 4 30 16 5 12 25 28i i i i i i i i= + − − − − + = + − + = +

c 1 2 12 1 2

i ii i

+ −+

+ −

2 2

2 2

1 2 2 1 1 22 2 1 2 1 2

2 3 2 1 24 1 4

4 3 35 5

7 45

7 45 5

i i i ii i i i

i i i ii i

i i

i

i

+ − − + = + + − − + + − + −

= + − − + +

= +

+=

= +

d ( )

2

2 2 2

1 3 1 3 1 3 3 4 3 13 12 9 13 9 133 4 3 4 25 254 4 9 16 252

i i i i i i i ii ii i ii

+ + + + + + − += = × = = = − +

− +− + −−

7 Given 5 44 3

i X iYi

−= +

+.

2

2

5 4 5 4 4 3 20 31 12 8 31 8 314 3 4 3 4 3 25 25 2516 9

i i i i i i ii i i i

− − − − + −= × = = = −

+ + − −

∴

8 3125 25

8 31and =25 25

i X iY

X Y

− = +

= −



8 a 2 2 2 2 2 2

1 1 x iy x iy x y iz x iy x iy x y x y x y

− −= × = = −

+ − + + +

b 2 2 2 2 2 2

1 1 x iy x iy x y iz x iy x iy x y x y x y

+ += × = = +

− + + + +

c 2

2 2 2 2 2

1 x y iz x y x y

= − + +

( )

( )

2 2 2

22 2

2 2

22 2

2 2

2 2 2 2 2 2

2

2

2( ) ( )

x xyi y i

x y

x y xyi

x y

x y xy ix y x y

− +=

+

− −=

+

−= −

+ +

d 2 2 2 2

1 1 x y iiz i x y x y= × −

+ +

2 2 2 2

2 2 2 2

2 2 2 2

1 i x y ii i x y x y

x yi ix y x yy x i

x y x y

= × × −+ +

= − × −+ +

= − −+ +

e z i x iy iz i x iy i− + −

=+ + +

( )( )

( )( )

( )( )

( )

( ) ( )

2 2 2

22 2

2 2

22

2 2

2 22 2

1 11 1

1 1 ( 1)

1

1 21

1 21 1

x i y x i yx i y x i y

x ix y y i y

x i y

x y ixx y

x y x ix y x y

+ − − += ×

+ + − +

+ − − − − −=

− +

+ − −=

+ +

+ −= −

+ + + +


9 If 1z i= + then 2 22 2 (1 ) 2(1 ) 2z z i i− + = + − + +

21 2 2 2 21 1 2 20

i i i= + + − − += − − +=

∴ 1z i= + is a solution of the equation 2 2 2 0z z− + = .

10 If 1 32iz − +

= then 2

2 1 3 1 31 12 2i iz z

− + − ++ + = + +

21 2 3 3 1 3 14 2

2 2 3 2 2 3 44 4 4

2 2 3

i i i

i i

i

− + − += + + − − − +

= + +

− −=

2 2 3i− + 44

040

+

=

=

∴1 3

2iz − +

= is a solution of the equation 2 1 0z z+ + = .


11 If 1 32iw − −

= then 3

3 1 32iw

− −=

( ) ( ) ( )

( )

( )

3

2 3

2 3

1 32

1 3 3 3 3 3using Pascal's triangle

8

1 3 3 9 3 3

8

1 3 3

i

i i i

i i i

i

+= −

+ + + = − + + + = −

+= −

9 3 3i− −

8

8

81

−= −

=

∴ 1 32iw − −

= is a solution of the equation 3 1w = .

12 a Show that ( )( ) 21 3 1 3 2 4x i x i x x− − − + = − + .

( )( ) ( ) ( )

( ) ( )22

2 2

2

2

1 3 1 3 1 3 1 3

1 3

2 1 32 1 32 4

x i x i x i x i

x i

x x ix xx x

− − − + = − − − +

= − −

= − + −

= − + +

= − +

b Show that ( )( ) 21 3 1 3 2 2 3x i x i x i+ − − + = + + .

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

22

2 2

2 2

2

1 3 1 3 1 3 1 3

1 3

1 3 3 3

1 3 3 3

2 3 3

x i x i x i x i

x i

x i i

x i i

x i

+ − − + = + − − −

= − −

= − − +

= − + +

= + +


Exercise 10.05: The complex plane


1 i a ( )1, 2A b ( )3, 1B − c ( )3, 4C − − d ( )2, 2D −

ii

2 3z i= + , 3 3u i= − + , 4 4v i= − − , 2 4w i= −

3

a

b


c

4 a 4z i= + , 2 2n i= − + , 5 2k i= − − , 2 3m i= −

b 4z i= − , 2 2n i= − − , 5 2k i= − + , 2 3m i= +

c


5 a ( )( ) 21 2 2 3z i i i i i= + − = + − = +

b 2

1 1 1 1 1 11 1 1 2 2 21

i i i iwi i i i

− − − − − −= = × = = = − −− + − + − − −

c ( )222 2 2 2 1 2 2v i i i i= − = − + = −


6 Given 3 2z i= +

a i w iz=

( )2

3 2

2 32 3

i i

i ii

= +

= += − +

ii 2v i z=

( )( )

2 3 2

1 3 23 2

i i

ii

= +

= − +

= − −

iii 3u i z=

( )( )

3

2

3 2

3 2

3 22 3

i i

i i

i ii

= +

= − +

= − −= −

b

c Multiplying a vector by i rotates the vector 90° anticlockwise



7 Given z = a + bi then 2a bi a biz z a+ + − =+ =

If 0z z+ = then a = 0

i.e. z = bi

which means that z lies on the imaginary axis.

8 Given z = a + bi then 2a bi a biz z a+ + − =+ =

If 0z z+ ≠ then a ≠ 0.

i.e. 2z z a+ = and as a ∈ R then z z+ lies on the real axis for z z+ ≠ 0 .

9 Given z = a + bi then if z z=

020

a bi a bibi b

bz

i

a

ib

+ = −= −=

⇒⇒⇒⇒ =∴ =

The position of z is on the real axis at a units from the origin.


Exercise 10.06: Modulus of a complex number


1 a 2 53 5 3 5 34i+ = + =

b ( )2 22 2 1 5i− + = − + =

c ( ) ( )2 23 3 1 4 2i− = + − = =

d ( )221 2 1 2 3i+ = + =

e 2 2

1 1 1 1 1 1 1 12 22 2 2 2

+ = + = + = =

i

f ( )223 4 1 53 4 15 5 5

i−= + − = =

2 Given 3 2u i= −

a ( )223 2 13u = + − =

b 1u−

( )

( )

12

1

1 3 2 3 2 1 3 23 2 3 2 139 4

1 1 13 13 2 9 413 13 13 13

i iu ii i i

u i

−

−

+ += × = = +

− + −

∴ = + = + = =

c 1 113u

=

d 2u

2u = (3 2i− )2 = 9 – 12i + 4i2 = 5 – 12i

∴ ( )22 25 12 13u = + − =


3 a ( )223 3 1 4 2i+ = + = =

b ( ) ( )2 22 7 2 7 2 7 3i− = + − = + =

c 1 16 8 10i

=+

d ( )221 1 1 11 1 1 2 12 2 2 2i i−= − = × + − = × =

4 a ( )5 2 3i i− + = 8 – 2i

∴ ( )5 2 3 8 2 68 2 17i i i− + = − = =

b ( )( )3 2 1 4 3 10 8 11 10i i i i− + = + + = +

∴ ( )( )3 2 1 4 11 10 221i i i− + = + =

c ( )2 3 2 4 3 2 3i i+ = +

∴ ( )2 3 2 4 3 2 3 48 12 60 2 15i i+ = + = + = =

d 2

1 1 1 3 1 3 1 31 3 41 3 1 3 1 3

i i iii i i

− − −= × = =

−+ + −

1 1 3 1 1 11 3 4

4 4 241 3i i

i−

∴ = = × − = × =+

e 1 1 1 1 2 11 1 1 2

i i i i ii i i

− − − − −= × = = −

+ + −

1 11

i ii

−∴ = − =

+

f ( ) ( ) ( )22 2 2 2 1 1 2 2 2 2i i i i i i i− = − − = − = +

( )22 2 2 8 1 3i i i∴ − = + = + =


5 Given 2 3z i= − + and 1 4w i= − , show that

a Show zw z w=

( )( )

( ) ( )

2

2 2

1 2

1 2

2 3 1 4

2 11 12

10 11

10 11

221

2 3 1 4

13 17

221

zw i i

i i

i

z z i i

z z

= − + −

= − + −

= +

= +

=

= − + −

=

=

=

b Show zz

w w=

2 31 4

z iw i

− +=

−

( ) ( )

2

2 2

1

2

2 3 1 41 4 1 4

2 5 1217

1 5 14171 14 5

1722117

2 31 4

13 1717 1722117

i ii i

i i

i

z iw i

zz

− + += ×

− +

− − +=

= − −

= − + −

=

− +=

−

= ×

=

=

11

2 2

zzz z

∴ =


c z w z w+ ≤ +

2 3 1 4

1

22 3 1 4

13 17

2 13 17

z w i i

i

z w i i

z w z w

+ = − + + −

= − −

=

+ = − + + −

= +

< +

∴ + ≤ +

d z w z w− ≥ −

( )2 3 1 4

3 7

587.6

2 3 1 4

13 170.5

13 17 58

z w i i

i

z w i i

z w z w

− = − + − −

= − +

=≈

− = − + − −

= −≈ −

− <

∴ − ≥ −

6 a x iy+ = 2 2x y+

b 2 2

1 1 + += × =

− − + +x iy x iy

x iy x iy x iy x y

2 2

2 2

2 22 2

2 2

2 2

1

1

1

+∴ =

− +

= × ++

= × ++

+=

+

x iyx iy x y

x iyx y

x yx y

x yx y


c 11

x iyx iy+ −+ +

( )( )

( )( )

( ) ( )( )

( ) ( )( )

( )( ) ( )

2 2

2 2 2

2 2

2 2

2 22 2

1 11 1

1 2 11

1 2 11

1 1 2 11

y

x iy x iyx iy x iy

x iy x i yx i y

x y iy xx y

x y iy xx y

+ − + −= ×

+ + + −

+ − + +=

+ −

+ − − +=

+ +

= × + − − + + +

∴

( )( ) ( )

( )( ) ( )

( )( ) ( ) ( )

( )( ) ( )

( )( )

( )( )

2 22 2

2 22 22 2

4 2 22 4 22 2

4 22 42 2

22 22 2

2 2

2 2

1 1 1 2 11 1

1 1 2 111 1 2 1 4 1

11 1 2 1

1

1 11

11

1

x iy x y iy xx iy x y

x y y xx y

x y x y y xx y

x y x yx y

x yx y

x yx y

+ −= × + − − +

+ + + +

= × + − + − + + +

= × + − + + + ++ +

= × + + + ++ +

= × + + + +

+ +=

+ +

=



7 z z= because the modulus is the length of the vector. The vectors z and z are

reflections so have the same length.

8 Show that 2 2z z= .

Let z x yi= +

( )( )

22 2 2 2 2

22

z x y x y

z x iy

= + = +

= +

( ) ( )

( )

2 2 2

2 2

2 22 2

4 2 2 4 2 2

4 2 2 4

22 2

2 2

2

2

2

2 4

2

x xyi i y

x y xyi

x y xy

x x y y x y

x x y y

x y

x y

= + +

= − +

= − +

= − + +

= + +

= +

= +

= 2z

∴ 2 2z z= .

9 If z x yi= + , show that 2z z z= .

( )22 2 2 2 2z x y x y= + = +

( )( ) 22 2 2 2 2z z x yi x yi x y i x y z= + − = − = + =


10 If z x yi= + and w u vi= + where , , ,u v w y∈R prove that

a Show zw z w=

( )( )( ) ( )

( ) ( )

( )( )

1 2

2 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2 2 2 2 2

2 2

z z x yi u vi

xu yv yu xv i

xu yv yu xv

x u xuyv y v y u xuyv x v

x u y u x v y v

z w x y u v

x y u v

x u y u x v y vzw

= + +

= − + +

= − + +

= − + + + +

= + + +

= + +

= + +

= + + +

=

1 2 1 2z z z z∴ =


b Show zz

w w=

( ) ( )

( ) ( )

( ) ( )

2 2

2 2

2 22 2

2 2 2 2 2 2 2 22 2

2 2 2 2 2 2 2 22 2

2 2 2 2 2 2

2 2 2 2 2 2

2

1

1

1 2 2

1

1

z x yiw u vi

x yi u viu vi u vixu yv yu xv i

u v

xu yv yu xv iu v

xu yv yu xvu v

x u xuyv y v y u xuyv x vu v

x u y u x v y vu v

z x yiw u vi

x y x y u vu v u v u v

u v

+=

+

+ −= ×

+ −

+ + −=

+

= + + −+

= + + −+

= − + + + ++

= + + +++

=+

+ + += = ×

+ + +

=+

( )( )

2 2 2 22

2 2 2 22 2

2 2 2 2 2 2 2 22 2

1

1

x y u v

x y u vu v

x u y u x v y vu vzw

+ +

= + ++

= + + ++

=

11

2 2

zzz z

∴ =


Exercise 10.07: Operations in the Argand plane


1

a

b

c

2

a


b

c

3 a

b

c


4

a

b

c


5 Consider the parallelogram OPQR in the Argand diagram below.

Let OP = u and OR = v.

a OQ = u + v

b PR = −v u

c QO = − −u v

d RP = −u v

e M, the midpoint of OQ.

OM = ( )12

u + v

6

OL = OM + ML

= OM + ON

= m + n

= 2 i− + + 3 2i+

OL = 1 + 3i



7 a 3 1 2= −z z z

b 1 2 3= +z z z

c 2 1 3= −z z z 8

a

b

a 1 2 1 2≤ +w + w w w

b 1 2 1 2≥ −w - w w w

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.


9

z

zz

zZ

Z

Z

Z

11

2

23

3

4

4u

w

v

Show that 4 1− = + +z z w u v .

( )( )

2 1

3 2

4 3

4 2

1

4 1

4 1

∴

∴ −

z = z + wz = z + vz = z + u

z = z + v + u

= z + w + v + u

z = z + w + v + u z z = w + v + u

,


Exercise 10.08: Properties of complex numbers


1 Given 1z a bi= + , 2z c di= + and 3z e fi= + .

Prove the following field properties for the set of Complex numbers :

a Closure under addition

1z + 2z = a bi+ + c di+ = (a+c) + (b+d)i

∴ closed under addition

b Closure under multiplication

1z 2z = ( )a bi+ ( )c di+ = (ac –bd) + (bc +ad)i

∴ closed under multiplication

c The commutative law of addition

1z + 2z = a bi+ + c di+ = c di+ + a bi+ = 2z + 1z

Complex numbers are commutative under addition.

d The commutative law of multiplication

1z 2z = ( )a bi+ ( )c di+ = ( )c di+ ( )a bi+ = 2z 1z

Complex numbers are commutative under multiplication.

e The associative law of addition

( 1z + 2z ) + 3z = ( a bi+ + c di+ ) +e fi+

= (a + c + bi + di) + e + fi

= (a + c + e) + (b + d + f)i

= a + (c + e) + bi + (d + f)i

= a + bi + (c + e + di + fi)

= a + bi + (c + di + e + fi)

= 1z + ( 2z + 3z )


f The associative law of multiplication

( 1z 2z ) 3z = {( a bi+ )( c di+ )}( e fi+ )

= {(ac –bd) + (bc +ad)i}( e fi+ )

= {(ace –bde) + (bce +ade)i}+{(acf – bdf)i – (bcf + adf)}

= (ace –bde – bcf – adf) + i(bce +ade + acf – bdf)

1z 2(z 3z ) = ( a bi+ ){( c di+ )( e fi+ )}

= ( a bi+ ){(ce – df ) + i(de + cf)}

= (ace – adf + bcei – bdfi + adei + acfi – bde – bcf)

= (ace – bde – bcf – adf) + i(bce +ade + acf – bdf)

= ( 1z 2z ) 3z

∴ 1z 2(z 3z ) = ( 1z 2z ) 3z

g The distributive law of multiplication over addition

1z ( 2z + 3z ) = ( a bi+ )( c di+ + e fi+ )

= ac + adi + ae + afi+ bci + bdi2 + bei + bfi2

= {(ac –bd)+ (bc+ ad)i}+ {(ae – bf) + (af + be)i}

= 1z 2z + 1z 3z

Complex numbers are closed using the distributive law of multiplication

over addition.


2 a The additive inverse of z x yi= + is x yi− −

b The multiplicative inverse of z x yi= + is 2 2

x yix y−+

( )( )

( ) ( )

2 2

11

x yi a bix yia bi

x yi x yix yi

x y

+ + =

−+ = ×

+ −

−=

+

c the additive identity for z x yi= + is 0 + 0i

d the multiplicative identity for is 1 0 .z x yi i= + +

3 Show that each of the following is true for 3 4z i= − .

a Show 2z z z=

2z = 52 = 25

z z = ( )( ) 223 4 3 4 9 16 25i i i z− + = − = =

b Show 2 2z z=

2z = 52 = 25

( ) 222 2 23 4 9 24 16 7 24 7 24 25z i i i z= − = − − = − − = + = =


4 Let 1 1 2z i= − and 2 1z i= − + .

a Show 1 2 1 2z z z z=

( )( )( ) ( )

2 21 2

1 2 1 2

1 2 1 1 3 1 3 10

1 2 1 5 2 10

z z i i i

z z i i z z

= − − + = + = + =

= − − + = = =

b Show 11

2 2

zzz z

=

( ) ( )2 21

2

1 1

2 2

1 2 1 2 1 1 2 1 1 103 3 11 1 1 2 2 2 2

1 2 5 2 101 22 2

z i i i i iz i i i

z i zz i z

− − − − − + −= = × = = − = − + =− + − + − −

−= = × = =− +

11

2 2

zzz z

∴ =

c Show 1 2 1 2z z z z+ = +

1 2

1 2 1 2

1 2 1

1 2 1 1 2 1

z z i i i i

z z i i i i i z z

+ = − − + = − =

+ = − + − + = + + − − = = +

1 2 1 2z z z z∴ + = +

d Show 1 2 1 2z z z z=

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

1 2

1 2 1 2

1 2 1 1 3 2 1 3 1 3

1 2 1 1 2 1 1 1 3 2 1 3

z z i i i i i

z z i i i i i i z z

= − − + = − + + = + = −

= − − + = + − − = − − − + = − =

1 2 1 2z z z z∴ =



5 Prove the following properties z∀ ∈C .

Let 1z a bi= + and 2z c di= +

a Show 2z = z z

( ) ( )( )( )

222 2 2 2 2

22 2 2 2 2

)z a bi a b a b

z z a bi a bi a b i a b z

= + = + = +

= + − = − = + =

∴ 2z = z z

b Show 2 2z z=

( )( )

( )

( ) ( )

( )

( )

( )

22 2 2 2 2

22

2 2 2

2 2

2 22 2

4 2 2 4 2

4 2 2 4

22 2

2 2

2

2

2

2 4

2

z a b a b

z a bi

a abi b i

a b abi

a b ab

a a b b a b

a a b b

a b

a b

= + = +

= +

= + +

= − +

= − +

= − + +

= + +

= +

= +


c Show 1 2 1 2z z z z=

( )( )( ) ( )

( ) ( )

( )( )

1 2

2 2

2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 21 2

2 2 2 2

2 2 2 2 2 2 2 2

1 2

2 2

z z a bi c di

ac bd bc ad i

ac bd bc ad

a c acbd b d b c acbd a d

a c b c a d b d

z z a b c d

a b c d

a c b c a d b dz z

= + +

= − + +

= − + +

= − + + + +

= + + +

= + +

= + +

= + + +

=

1 2 1 2z z z z∴ =


d Show 11

2 2

zzz z

=

( ) ( )

( ) ( )

( ) ( )

1

2

2 2

2 2

2 22 2

2 2 2 2 2 2 22 2

2 2 2 2 2 2 2 22 2

1

2

2 2 2 2 2 2

2 2 2 2 2 2

1

1

1 2 2

1

1

z a biz c di

a bi c dic di c diac bd bc ad i

c d

ac bd bc ad ic d

ac bd bc adc d

a c acbd b d b c acbd a dc d

a c b c a d b dc d

z a biz c di

a b a b c dc d c d c d

c

+=

+

+ −= ×

+ −

+ + −=

+

= + + −+

= + + −+

= − + + + ++

= + + +++

=+

+ + += = ×

+ + +

=

( )( )

2 2 2 22 2

2 2 2 22 2

2 2 2 2 2 2 2 22 2

1

2

1

1

a b c dd

a b c dc d

a c b c a d b dc dzz

+ ++

= + ++

= + + ++

=

11

2 2

zzz z

∴ =


e Show 1 2 1 2z z z z+ = +

( ) ( )( ) ( )

1 2

1 2

z z a bi c di

a c b d i

a c b d ia bi c di

z z

+ = + + +

= + + +

= + − +

= − + −

= +

1 2 1 2z z z z∴ + = +

f Show 1 2 1 2z z z z=

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

( )

1 2

1 2

2

1 2

( )

( )

( )

z z a bi c di

ac bd bc ad i

ac bd bc ad i

z z a bi c di

a bi c di

ac bdi adi bciac bd bc ad i

z z

= + +

= − + +

= − − +

= + +

= − −

= + − −

= − − +

=

1 2 1 2z z z z∴ =

6 Show z z+ is real z∀ ∈C .

Let z x yi= +

z z+ = 2x yi x yi x+ + − =

which is real z∀ ∈C (assuming that x and y are real).

7 Show that z z is real z∀ ∈C .

Let z x yi= +

z z = ( )( ) 2 2 2 2x yi x yi x y i x y+ − = − = +

which is real z∀ ∈C (assuming x and y are real)


8 Let z x yi= +

( )

( )

2 2 2 2 22 2

2 2

1 1 1 1

1 1

x yz x yi x y

z x yi

= = =++ +

=+

( )

( ) ( )

( )

2 2

2 2

2 22 2

4 2 2 4 2 2

4 2 2 4

22 2

2 2

12

12

1

2

1

2 41

21

1

ix xyi y i

x y xyi

x y xy

x x y y x y

x x y y

x y

x y

=+ +

=− +

=− +

=− + +

=+ +

=+

=+

∴ 2 2

1 1zz

= z∀ ∈C .


Exercise 10.09: Quadratic equations


1 a 2 2 4 0z z+ + =

2

2

2

4 16 12 12

42

2 122

2 2 32

1 3

i

b b acxa

i

ix

x i

∆ = − = − =

− ± −=

− ±=

− ±=

= − ±

b 2 2 4 0z z− − =

2

4 16 20

42

2 202

2 2 52

1 5

b b acxa

x

x

∆ = + =

− ± −=

±=

±=

= ±

c 2 4 8 0z z+ + =

2

2

2

16 32 16 16

42

4 162

4 42

2 2

i

b b acxa

i

ix

x i

∆ = − = − =

− ± −=

− ±=

− ±=

= − ±


d 23 2 8 0z z+ + =

2

2

2

4 96 92 92

42

2 926

2 2 236

1 233

i

b b acxa

i

ix

ix

∆ = − = − =

− ± −=

− ±=

− ±=

− ±=

2 ( )21 9 0x + + =

( )21x + = –9

( )21x + = 9i2

x + 1 = ± 3i

1 3x i= − ±

3 ( ) ( )2 21 1 2 3x x+ = − +

2 2

2

2 1 2 12 18 1 03 14 18 0x x x xx x+ + + + + − =

+ + =

2 2

2

2

14 216 20 20

42

14 206

14 2 56

7 53

i

b b acxa

i

ix

ix

∆ = − = − =

− ± −=

− ±=

− ±=

− ±=


4 Check that 5 iβ = − − is a root of the real quadratic equation 2 10 26 0z z+ + = .

2 10 26z z+ + = ( 5 i− − )2 + 10( 5 i− − ) + 26

= 25 + 10i + i2 –50 – 10i + 26

= 51 – 50 + i2

= 1 – 1

= 0 so 5 iβ = − − is a root of the real quadratic equation.

The other root is –5+i (all coefficients are real so the roots form a conjugate pair).

5 2 5 0z mz+ + =

2 i− is a solution so

( ) ( )( )( )

( )

2

2

2 2 5 0

4 4 2 5 0

9 1 4 2 0

2 8 48 42

242

4

i m i

i i m i

i m i

m i iim

iimi

m

− + − + =

− + + − + =

− − + − =

− = − +

− +=

−− = − −

= −

All coefficients are real so the other root is 2 + i.

6 Form quadratic equations for which the following complex conjugates are roots. Give your answer in the form

i ( )( ) 0x x− α − α = ii ( )2 0x x− α + α + αα = .

a 1 2i± i ( )( )1 2 1 2 0x i x i− + − − =

ii

2 1 2x i− + 1 2i+ −( ) ( )( )2 2

2

1 2 1 2 0

2 1 4 02 5 0

x i i

x x ix x

+ + − =

− + − =

− + =


b 3 i± i ( )( )3 3 0x i x i− − − + =

ii

( ) ( )( )2

2 2

2

3 3 3 3 0

2 3 3 0

2 3 4 0

x i i x i i

x x i

x x

− + + − + − + =

− + − =

− + =

c 4 2i− ± i ( )( )4 2 4 2 0x i x i+ + + − =

ii

2 4 2x i− − + 4 2i− −( ) ( )( )2 2

2

4 2 4 2 0

8 16 2 08 18 0

x i i

x x ix x

+ − + − − =

+ + − =

+ + =

d 1 12 2

i± i 1 1 1 1 02 2 2 2

x i x i − − − + =

ii

2

2 2

2

1 1 1 1 1 1 1 1 02 2 2 2 2 2 2 2

1 12 02 2

2 1 0

x i i x i i

x x i

x x

− + + − + + − =

− + − =

− + =

e 5 6i− ± i ( )( )5 6 5 6 0x i x i+ − + + =

ii

( ) ( )( )2

2 2

2

5 6 5 6 5 6 5 6 0

10 25 36 010 61 0

x i i x i i

x x ix x

− − + − − + − + − − =

+ + − =

+ + =

f 32 2

i± i

3 3 02 2 2 2

i ix x

− − − + =

ii

2

2 2

2

3 3 3 3 02 2 2 2 2 2 2 2

3 13 04 4

3 1 0

i i i ix x

x x i

x x

− − − − + + + − =

− + − =

− + =


g 3 25 5

i− ± i 3 2 3 2 05 5 5 5

i ix x + − + + =

ii

2

22

2

3 2 3 2 3 2 3 2 05 5 5 5 5 5 5 5

6 9 4 05 25 25

6 13 05 25

i i i ix x

x ix

xx

− − + + + − + =

− + − =

− + =

h 1 23 3

i− ± i 1 2 1 2 03 3 3 3

i ix x

+ − + + =

ii

2

2 2

2

1 2 1 2 1 2 1 2 03 3 3 3 3 3 3 3

2 1 2 03 33

2 1 03

x i i x i i

xx i

xx

− − − − + + − − − + =

+ + − =

+ + =


7 2 0kz nz p+ + =

a 2 i±

2 iα = + and 2 iα = −

( ) ( )

( )( )

2

2 24

2 254 5 0

1, 4 and 5

i i

i i

x xk n p

+ = + + −

=

= + −

=

∴ − + == = − =

α α

αα

b 1 3i±

1 3iα = + and 1 3iα = −

( )( )

2

1 3 1 32

1 3 1 3

42 4 0

1, 2 and 4

i i

i i

x xk n p

+ = + + −=

= + −

=

∴ − + == = − =

α α

αα


c 1 154i− ±

1 154iα − +

= and 1 154iα − −

=

2

2

2

1 15 1 154 4

12

1 15 1 154 4

1 1516

1

1 02

. .2 2 0

2, 1 and 2

i i

i i

i

xx

i ex x

k n p

− + − −+ = +

= −

− + − −= −

=

=

∴ + + =

+ + == = =

α α

αα

d 5 1213 13

i±

5 1213 13

iα = + and 5 1213 13

iα = −

2

2

2

5 12 5 1213 13 13 1310135 12 5 12

13 13 13 1325 144

169 1691

10 1 013

. .13 10 13 0

13, 10 and 13

i i

i i

i

xx

i ex x

k n p

+ = + + −

=

= + −

= −

=

∴ − + =

− + == = − =

α α

αα


8 a 2 2 2z z− + = 2 2 22 1 1 ( 1) ( 1 )( 1 )z z z i z i z i− + + = + − = + − + +

b ( )( )2 2 2 22 6 2 1 5 ( 1) 5 1 5 1 5z z z z z i z i z i− + = − + + = + − = + − + +

c 2 2 2 24 5 4 4 1 ( 2) ( 2 )( 2 )z z z z z i z i z i+ + = + + + = + − = + − + +

d 2 2

2 2 1 3 1 3 1 3 1 314 4 2 4 2 2 2 2

i i iz z z z z z z + + = + + + = + − = + − + +

.


9 3 1z =

⇒3

2

1 0( 1)( 1) 0zz z z− =

− + + =

z = 1 or 2( 1) 0z z+ + =

2

2

2

1 4 3 3

42

1 32

1 32

1 3 1 3or z =2 2

i

b b acza

i

iz

i iz

∆ = − = − =

− ± −=

− ±=

− ±=

− + − −=

The complex roots are conjugates as required. z = 1 is the real root.

10 Let ω = 1 32i− + and ω = 1 3

2i− − (See question 9)

a Show 2ω = ω

( )2 2

22 1 31 3 1 2 3 3 1 2 3 3 1 3

2 4 4 4 2

ii i i i i− − − + − + − − − −ω = = = = = = ω


b Show 2 1ω = ω = ω =

1 3 1 4 1

2 2i− +

ω = = =

1 3 1 11 3 4 12 2 2i i− −

ω = = + = =

2

2 1 3 1 3 12 2i i − + − −

ω = = = ω =

∴ 2 1ω = ω = ω =

11 The solutions of 3 1z = are z = 1, 1 3 1 3and z =2 2i iz − + − −

= ’

The three solutions are symmetrically around O. i.e. the complex numbers are plotted 120° apart. The magnitude of each solution is 1.


Chapter 10 Review

Multiple choice

1 2 1 3 4 2 5z w i i i+ = + + − = + C

2 ( )( )1 3 2 2 7 3 1 7zw i i i i= − − = − − = − − A

3 ( )26107 4 3 21i i i i i i= × = × × = − B

4 1 1 1 1 1 1 1

1 22 2 22 2z ii= = × = × =

++ D

5 zw z w≠

is always a real numberis probably a complex number

zwz w

C

Short answer

6 a 281 81 9i i− = =

b 218 18 3 2i i− = =

c 248 48 4 3i i− = =

d 2216 216 6 6i i− = =

7 a 2 236 36

6z iz i

= − == ±

b 2 9 0z + =

2

2 2

993

zz iz i

= −

== ±

c 24 1 0+ =z

2

22

14

4

2

= −

=

= ±

z

iz

iz


8 a 3 2i i i i= = −

b ( ) ( )100 100200 2 1 1i i= = − =

c ( ) ( )24 2449 48 2 1i i i i i i i= = = − =

d ( ) ( ) ( )295 294590 2 1 1 1i i= = − − = −

e ( ) ( )17 17 18 9 92

1 11

i i i i iii i i i

= × = = = = −−

9 a Re( 5 5i− ) = 5 Im( 5 5i− ) = 5−

b Re1 3 1

2 2i +

=

Im1 3 3

2 2i +

=

c 3 2 ( 2 ) (3 )x yi y xi x y i y x+ − + = − + +

Re [ ]( 2 ) (3 )x y i y x− + + = x – 2y

Im[ ]( 2 ) (3 )x y i y x− + + = 3y + x

10 a 2 4 6 0x x+ + = 16 24 8∆ = − = −

2

2

42

4 82

4 82

4 2 22

2 2

b b acxa

i

i

x i

− ± −=

− ± −=

− ±=

− ±=

= − ±


b 23 10 12 0z z− + = 100 144 44∆ = − = −

2

2

42

10 446

10 446

10 2 116

5 113

b b acxa

i

i

ix

− ± −=

± −=

±=

±=

±=

c 2 2 0w w+ + = 1 8 7∆ = − = −

2

2

42

1 72

1 72

1 2 72

1 72

b b acxa

i

i

ix

− ± −=

− ± −=

− ±=

− ±=

− ±=

11 a 4 7 4 7i i− + = − −

b 2 5 2 5i i+ = −

c 2 3 2 3x y yi xi x y yi xi+ − + = + + −

12 a 2 2 2 24 5 4 4 1 ( 2) ( 2 )( 2 )+ + = + + + = + − = + − + +x x x x x i x i x i

b 2 2 2 210 29 10 25 4 ( 5) 4 ( 5 2 )( 5 2 )z z z z z i z i z i− + = − + + = − − = − − − +

c 2 2

2 2 1 3 1 3 1 3 1 314 4 2 4 2 2 2 2

iw w w w w w i w i + + = + + + = + − = + − + +


13 a 3 i±

3 iα = + and 3 iα = −

( )( )

2

2

3 36

3 3

910

6 10 0

i i

i i

i

x x

α + α = + + −=

αα = + −

= −=

∴ − + =

b 1 6i±

1 6iα = + and 1 6iα = −

( )( )2

2

1 6 1 62

1 6 1 6

1 672 7 0

i i

i i

i

x x

α + α = + + −=

αα = + −

= −=

∴ − + =

c 2 32i±

2 32i+

α = and 2 32i−

α =

2

2

2

2 3 2 32 2

2

2 3 2 32 2

4 34

74

72 04

4 8 7 0

i i

i i

i

x x

x x

+ −α + α = +

=

+ −αα =

−

=

=

∴ − + =

∴ − + =


14 2 ( 2 ) 4 7x y i x y i+ + − = +

2 4 and 2 72(4 2 ) 4

5 122.44 2(2.4)

0.8

x y x yx xx

xyy

∴ + = − =− − ==== −= −

x = 2.4 and y = –0.8

15 a 2 8 4 6 2 2i i i+ − − = − +

b ( )( ) 24 7 3 28 5 3 31 5i i i i i− + = + − = +

c ( )2 23 2 9 12 4 5 12i i i i− = − + = −

16 a 2

2

1 1 2 2 3 1 3 1 32 2 2 5 5 54

i i i i i i ii i i i

+ + + + + += × = = = +

− − + −

b 2

1 1 2 2 2 23 3 322 2 2

i i i iii i i

− − −= × = = = −

−+ + −

c ( )( )( ) ( )( ) 2

2 1 (1 )2 2 4 4 4 21 1 1 1 1 1 21

i i i i i ii i i i i i i

− − + − − −− = = = = = −

+ − + − + − −

17 a 2z i= + b 3 2w i= − − c 4 4 4 41

i iv ii i i

= = × = = −−


18

19

20 a 2 3z i= − +

4 9 13z = + =

b 1 34iz +

=

1 3 1 116 16 4 2

z = + = =

c 2

1 1 3 2 3 2 3 273 43 2 3 2 3 2

i i izii i i

+ + += = × = =

−− − +

1 1 73 2 77 7 7

z i= + = =


21 ( ) ( )2x y i x y− − +

( ) ( )2 2

2 2 2 2

2 2

2

2 4 4

5 2 2

x y x y

x xy y x xy y

x xy y

= − + − −

= − + + + +

= + +

22

23 2 4z i= −

a 12

z = 1 – 2i 2 4z z i− = − +

so U(1, –2) and V(–2, 4)

b


24

25 a The multiplicative inverse of 1 2i− is a + bi

such that ( ) ( )1 2 1i a bi− × + =

( ) ( )

( )( )( )

2

11 2

1 211 2 1 2

1 21 2

1 23

a bii

i

i i

ii

i

∴ + =−

+= ×

− +

+=

−+

=

b The multiplicative inverse of 32i is a + bi

such that ( )32 1i a bi× + =

( ) 3

121212

2

∴ + =

= −

= − ×

= −

a bii

ii

i ii


c The multiplicative inverse of 13 4i−

is a + bi

such that ( )1 13 4

a bii× + =

−

( ) 3 4∴ + = −a bi i

26 Given z a bi= + and w c di= +

a Show z w z w+ = +

( )

( )

z w a bi c dia c i b d

z w a c i b da bi c diz w

z w z w

+ = + + += + + +

∴ + = + − += − + −= +

∴ + = +

b Show zw z w=

( )( )

( )( )

( )

( )

(

zw a bi c diab bd i bc ad

zw ab bd i bc ad

z w a bi c diac bd i bc ad

zw

zw z w

= + +

= − + +

∴ = − − +

= − −

= − − +

=

∴ =

27 a 2 4 0x + =

2

2

442

xi

x i

= −

== ±

Imaginary roots


b 23 10 3 0z z− + =

100 36 64∆ = − =

2 42

10 646

10 86

13 or3

b b acza

z z

− ± −=

±=

±=

= =

Real roots c 2 2 2 0w w− + =

4 8 4∆ = − = −

2

2

42

2 42

2 22

1

b b acwa

i

i

w i

− ± −=

±=

±=

= ±

Complex roots

28 ( ) ( )22 8 17 4 8 4 17z w i i− + = − − − +

216 8 32 8 1733 32 1 00

4 is a root

i i ii

i

= − + − + += − − +=

∴ −

The other root is 4 + i


29 3 2i− ±

3 2iα = − + and 3 2iα = − −

( ) ( )3 2 3 2x i x i − − + − − − = 0

( )( )3 2 3 2 0z i z i+ − + + =

30 a ( )( )5 2 6 7 30 23 14 44 23i i i i− + = + + = +

b 6 8 36 64 10i− = + =

c ( )2 2 2 2 2 2 1 1 1 2 22 1 32 2 2

i i i i ii i i

+ + + + −= × = = +

+− − +

d 3 4 3 4i i− = + 31 Given the complex numbers 3u i= − and 1 3v i= − − .

a u v+ ( ) ( )3 1 3 3 1 1 3= − − − = − − +i i i

b ( ) ( ) ( )Re Re 3 1 3 Re 3 1 3 3 1u v i i i i+ = − − − = − − − = −

c ( ) ( )Im Im 3 1 3 1 3u v i i+ = − − − = − −

d ( ) ( )3 1 1 3u v i+ = − + − −

( ) ( )2 23 1 1 3

3 2 3

= − + − −

= − 1 1 2 3+ + + 3

8

2 2

+

=

=

32

a


b

c

d

33 a 2 25 0z + = 2 225 25

5z iz i

= − == ±

b 2 10 34 0z z− + = 100 136 36∆ = − = −

2

2

42

10 362

10 362

10 62

5 3

b b acza

i

i

z i

− ± −=

± −=

±=

±=

= ±


c 2 3 0 ( 3) 0w w w w+ = + =

0 or 3w w= = −

d 22 8 9 0u u+ + = 64 72 8∆ = − = −

2

2

42

8 84

8 84

8 2 24

222

b b acua

i

i

u i

− ± −=

− ± −=

− ±=

− ±=

= − ±

e 23 7 5 0z z+ + = 49 60 11∆ = − = −

2

2

42

7 116

7 116

7 116

7 116 6

b b acza

i

i

z i

− ± −=

− ± −=

− ±=

− ±=

= − ±

f 4 25 36 0w w− − =

( )( )2 2

2 2

9 4 0

3 or 4 43 or 2

− + =

= ± = − == ± =±

w w

w w iw w i

34 Roots are 2 2i− ±

( )( )

2

4 2 2 2 2 4 4 8

4 8 0

i i

x x

+ = − = − + − − = + =

+ + =

α α αα


Application

35 2 5 3x x+ = 2 3 5 0x x− + = 9 20 11∆ = − = −

2

2

42

3 112

3 112

3 112

3 112 2

b b acza

i

i

z i

− ± −=

± −=

±=

±=

= − ±

36 3 8z =

( )( )

3

2

8 0

2 2 4 0

z

z z z

− =

− + + =

Either z = 2 or ( )2 2 4 0z z+ + = 4 16 12∆ = − = −

2

2

42

2 122

2 122

2 2 32

1 3

b b acza

i

i

z i

− ± −=

− ± −=

− ±=

− ±=

= − ±


37 VWXU is a parallelogram.

V(1,1)

W(4, 2)

x-2 -1 1 2 3 4 5

y

-2

-1

1

2

3

4

5

U(-1, 3)

X

O

M

OX = OU + UX UX = VW = 3i + j

OX = – i + 3j + 3i + j

= 2i + 4j

The coordinates of X are (2, 4)

The midpoint M of VX = ?

OM = OV + 0.5 VX

= (1, 1) + 0.5(1, 3)

= (1.5, 2.5)

∴M(1.5, 2.5)


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