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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
© Cengage Learning Australia 2014 ISBN 9780170251501 2
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
© Cengage Learning Australia 2014 ISBN 9780170251501 3
Exercise 10.02: Complex numbers
Concepts and techniques
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+
=+
© Cengage Learning Australia 2014 ISBN 9780170251501 4
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
±=
±=
= + = −
© Cengage Learning Australia 2014 ISBN 9780170251501 5
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
±=
±=
= + = −
© Cengage Learning Australia 2014 ISBN 9780170251501 6
Reasoning and communication
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
© Cengage Learning Australia 2014 ISBN 9780170251501 7
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
∆ = − + = −
− ± −=
±=
±=
= ±
© Cengage Learning Australia 2014 ISBN 9780170251501 8
Exercise 10.03: Operations with Complex numbers
Concepts and techniques
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)
© Cengage Learning Australia 2014 ISBN 9780170251501 9
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
∆ = − = − =
− ± −=
− ±=
− ±=
= − + = − −
© Cengage Learning Australia 2014 ISBN 9780170251501 10
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
− + + − =
− + − =
− + =
© Cengage Learning Australia 2014 ISBN 9780170251501 11
Reasoning and communication
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
∆ = − − = − =
− ± −=
±=
±=
±= ×
±=
− − ±
=
= −
© Cengage Learning Australia 2014 ISBN 9780170251501 12
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=
© Cengage Learning Australia 2014 ISBN 9780170251501 13
Exercise 10.04: Operations with complex numbers
Concepts and techniques
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+ + − = + + − = −
© Cengage Learning Australia 2014 ISBN 9780170251501 14
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
− + − + − += = = × = = = − +
− − − +− + −− +
© Cengage Learning Australia 2014 ISBN 9780170251501 15
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
− = +
= −
© Cengage Learning Australia 2014 ISBN 9780170251501 16
Reasoning and communication
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
+ − − += ×
+ + − +
+ − − − − −=
− +
+ − −=
+ +
+ −= −
+ + + +
© Cengage Learning Australia 2014 ISBN 9780170251501 17
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+ + = .
© Cengage Learning Australia 2014 ISBN 9780170251501 18
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
+ − − + = + − − −
= − −
= − − +
= − + +
= + +
© Cengage Learning Australia 2014 ISBN 9780170251501 19
Exercise 10.05: The complex plane
Concepts and techniques
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
© Cengage Learning Australia 2014 ISBN 9780170251501 20
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
© Cengage Learning Australia 2014 ISBN 9780170251501 21
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= − = − + = −
© Cengage Learning Australia 2014 ISBN 9780170251501 22
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
© Cengage Learning Australia 2014 ISBN 9780170251501 23
Reasoning and communication
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.
© Cengage Learning Australia 2014 ISBN 9780170251501 24
Exercise 10.06: Modulus of a complex number
Concepts and techniques
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 = + − =
© Cengage Learning Australia 2014 ISBN 9780170251501 25
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∴ − = + = + =
© Cengage Learning Australia 2014 ISBN 9780170251501 26
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
∴ =
© Cengage Learning Australia 2014 ISBN 9780170251501 27
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
© Cengage Learning Australia 2014 ISBN 9780170251501 28
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
+ −= × + − − +
+ + + +
= × + − + − + + +
= × + − + + + ++ +
= × + + + ++ +
= × + + + +
+ +=
+ +
=
© Cengage Learning Australia 2014 ISBN 9780170251501 29
Reasoning and communication
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= + − = − = + =
© Cengage Learning Australia 2014 ISBN 9780170251501 30
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∴ =
© Cengage Learning Australia 2014 ISBN 9780170251501 31
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
∴ =
© Cengage Learning Australia 2014 ISBN 9780170251501 32
Exercise 10.07: Operations in the Argand plane
Concepts and techniques
1
a
b
c
2
a
© Cengage Learning Australia 2014 ISBN 9780170251501 33
b
c
3 a
b
c
© Cengage Learning Australia 2014 ISBN 9780170251501 34
4
a
b
c
© Cengage Learning Australia 2014 ISBN 9780170251501 35
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
© Cengage Learning Australia 2014 ISBN 9780170251501 36
Reasoning and communication
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.
© Cengage Learning Australia 2014 ISBN 9780170251501 37
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
,
© Cengage Learning Australia 2014 ISBN 9780170251501 38
Exercise 10.08: Properties of complex numbers
Concepts and techniques
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 )
© Cengage Learning Australia 2014 ISBN 9780170251501 39
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.
© Cengage Learning Australia 2014 ISBN 9780170251501 40
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= − = − − = − − = + = =
© Cengage Learning Australia 2014 ISBN 9780170251501 41
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∴ =
© Cengage Learning Australia 2014 ISBN 9780170251501 42
Reasoning and communication
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
= + = +
= +
= + +
= − +
= − +
= − + +
= + +
= +
= +
© Cengage Learning Australia 2014 ISBN 9780170251501 43
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∴ =
© Cengage Learning Australia 2014 ISBN 9780170251501 44
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
∴ =
© Cengage Learning Australia 2014 ISBN 9780170251501 45
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)
© Cengage Learning Australia 2014 ISBN 9780170251501 46
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 .
© Cengage Learning Australia 2014 ISBN 9780170251501 47
Exercise 10.09: Quadratic equations
Concepts and techniques
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
∆ = − = − =
− ± −=
− ±=
− ±=
= − ±
© Cengage Learning Australia 2014 ISBN 9780170251501 48
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
∆ = − = − =
− ± −=
− ±=
− ±=
− ±=
© Cengage Learning Australia 2014 ISBN 9780170251501 49
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
+ + − =
− + − =
− + =
© Cengage Learning Australia 2014 ISBN 9780170251501 50
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
− − − − + + + − =
− + − =
− + =
© Cengage Learning Australia 2014 ISBN 9780170251501 51
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
− − − − + + − − − + =
+ + − =
+ + =
© Cengage Learning Australia 2014 ISBN 9780170251501 52
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
© Cengage Learning Australia 2014 ISBN 9780170251501 53
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
© Cengage Learning Australia 2014 ISBN 9780170251501 54
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 + + = + + + = + − = + − + +
.
Reasoning and communication
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
© Cengage Learning Australia 2014 ISBN 9780170251501 55
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.
© Cengage Learning Australia 2014 ISBN 9780170251501 56
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
© Cengage Learning Australia 2014 ISBN 9780170251501 57
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
− ± −=
− ± −=
− ±=
− ±=
= − ±
© Cengage Learning Australia 2014 ISBN 9780170251501 58
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 + + = + + + = + − = + − + +
© Cengage Learning Australia 2014 ISBN 9780170251501 59
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 =
−
=
=
∴ − + =
∴ − + =
© Cengage Learning Australia 2014 ISBN 9780170251501 60
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
= = × = = −−
© Cengage Learning Australia 2014 ISBN 9780170251501 61
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= + = =
© Cengage Learning Australia 2014 ISBN 9780170251501 62
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
© Cengage Learning Australia 2014 ISBN 9780170251501 63
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
© Cengage Learning Australia 2014 ISBN 9780170251501 64
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
© Cengage Learning Australia 2014 ISBN 9780170251501 65
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
© Cengage Learning Australia 2014 ISBN 9780170251501 66
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
© Cengage Learning Australia 2014 ISBN 9780170251501 67
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
− ± −=
± −=
±=
±=
= ±
© Cengage Learning Australia 2014 ISBN 9780170251501 68
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
© Cengage Learning Australia 2014 ISBN 9780170251501 69
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
− ± −=
− ± −=
− ±=
− ±=
= − ±
© Cengage Learning Australia 2014 ISBN 9780170251501 70
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)
© Cengage Learning Australia 2014 ISBN 9780170251501 71