Chapter 4

© Oxford Fajar Sdn. Bhd. (008974-T) 2013

cHAPTER 4 comPlEx numbERs

Focus on Exam 4

1 1 + 2i1 – i

= 1 + 2i1 – i

1 + i1 + i

= 1 + 2(i)2 + 2i + i1 – i2

= 1 – 2 + 3i1 + 1

= –1 + 3i

2

[x = – 12

and y = 32

2 z1* = 3 + i, z2* = 1 – i

1z*1

+ z1z2* = 13 + i

+ (3 – i)(1 – i)

= 1(3 + i)

(3 – i)(3 – i)

+ 3 – 1 – 4i

= 3 – i9 + 1

+ 2 – 4i

= 310

+ 2 – 110

i – 4i

= 2310

– 4110

i

[ a = 2310

, b = – 4110

3 z1* = 1 – 5i, z2* = 2 – iz1z2* + z1*z2 = (1 + 5i)(2 – i) + (1 – 5i)(2 + i)

= 2 + 5 + 9i + 2 + 5 – 9i= 14 (Shown)

4 (a) z = 2 – 2i ⇒ z* = 2 + 2i z + z* = 2 – 2i + 2 + 2i

= 4 z – z* = 2 – 2i – (2 + 2i)

= – 4i= –i(z + z*) (Shown)

(b) 1z + 1

z* = 1

2 – 2i + 1

2 + 2i

= 1(2 – 2i)

ii +

1(2 + 2i)

ii

= 12 + 2

+ i2i – 2

= i2 – 2i

– i2 – 2i

= i1 1z*

– 1z2 [Shown]

5 3 – ai

1 – 3 i =

3 – ai1 – 3 i

1 + 3 i1 + 3 i

= 3 + a 3 + (3 – a)i

1 + 3

= 3 (a + 1) + (3 – a)i

4

3 – ai

1 – 3 i is a real number ⇒ 3 – a = 0

a = 3

Hence the real number is 3 (3 + 1)

4 = 3

6 (x + iy)2 = 4i x2 – y2 + 2xyi = 4i Comparing the real and imaginary parts: x2 – y2 = 0 …j 2xy = 4

y = 2x …k

Substitute y = 2x into j:

x2 – 12x2

2

= 0

x4 – 4 = 0 (x2 + 2)(x2 – 2) = 0


Fully Worked Solution 3ACE AHEAD Mathematics (T) First Term Updated Edition2

x2 = –2 or x2 = 2 [ x = ± 2 (reject x2 = –2)

Substitute x = ± 2 into k:

y = ± 22

= ± 2 Hence x = ± 2, y = ± 2

7 Let z = a + ib ⇒ z* = a – ib (a + ib)(a – ib) – 5i(a + ib) = 10 – 20i a2 + b2 – 5ai + 5b = 10 – 20i (a2 + b2 + 5b) – 5ai = 10 – 20i Equating the real and imaginary parts: ⇒ a2 + b2 + 5b = 10 …j

⇒ 5a = 20a = 4

Substitute a = 4 into j:16 + b2 + 5b = 10b2 + 5b + 6 = 0

(b + 2)(b + 3) = 0b = –2 or b = –3

[ z = 4 – 2i or z = 4 – 3i

8 z = x + iyiz = i(x + iy) = –y + xiz + iz = (x + iy) + (–y + xi)

= (x – y) + (x + y)i

Imaginary

Real–y x

y

x

x + y

O

R(x − y, x + y)

P(x, y)

Q (−y, x )

OP is perpendicular to OQ ⇒ POQ = 90°⇒ OPRQ is a squarePoint R: z + iz = (x – y) + (x + y)iWhen x = y, z + iz = (y – y) + (y + y)i

= 2yi⇒ R lies on the imaginary axis.

9 z4 – 2z3 + kz2 – 18z + 45 = 0If z = ai (a is a constant) is a root, then z = –ai is also a root.⇒ (z – ai)(z + ai)(z2 + bz + c)

= z4 – 2z3 + kz2 – 18z + 45(z2 + a2) + (z2 + bz + c) = z4 – 2z3 + kz3 + 18z + 45

z4 + bz3 + (a2 + c)z2 + a2bz + ca2 = z4 – 2z3 + kz2 – 18z + 45

Comparing the coefficients of z3: b = –2Comparing the coefficients of z: a2b = –18

⇒a2 = 9Comparing the constants: ca2 = 45

⇒c = 5∴ z2 + b2z + c = z2 – 2z + 5 = 0

z = 2 ± 4 – 4(1)(5)

2

= 2 ± –16

2= 1 ± 2i

Hence the roots are ±3i, 1 ± 2iComparing the terms (a2 + c)z2 and kz2

k = a2 + c= 9 + 5= 14

10 (a) z3 = z1z2

= (4 – 3i)(2 + i)= 8 + 3 – 2i= 11 – 2i

|z3| = 112 + (–2)2

= 11.18 (b) arg z3 = tan–1 1–2

112= –tan–1 1 2

112= –0.18 radian

(c) Imaginary

RealO

(11, –2)z3

11 z1z2 = (1 – 3 i)( 3 + i)

= 3 + 3 + (1 – 3)i

= 2 3 – 2i



r = |z1z2| = (2 3 )2 + (–2)2

= 12 + 4= 4

q = –tan–1 1 22 3 2

= – p6

[z1z2 = 43cos 1– p62 + i sin 1–

p624

Imaginary

RealO

r

(2 3, –2)

q

12 Let z = a – ib (a – ib)2 = 1 – 2 2 i a2 – b2 – 2abi = 1 – 2 2 i Equating the real and imaginary parts:

a2 – b2 = 1 …j2ab = 2 2

b = 2

a …k

Substitute b = 2

a into j:

a2 – 2a2

= 1

a4 – a2 – 2 = 0(a2 – 2)(a2 + 1) = 0

a2 = 2 (reject a2 = –1)a = ± 2

Substitute a = ± 2 into k ⇒ b = ± 2 2

= 1 Hence z1 = 2 – i and z2 = – 2 + i

(b) Imaginary

RealO

( 2, –1)

(– 2, 1)

z2

z1

(c) |z1| = ( 2)2 + (–1)2 = 3

|z2| = (– 2)2 + (1)2 = 3

arg z1 = –tan–1 1 12 2 = –0.615 radian

arg z2 = p– tan–1 1 12 2 = 2.526 radian

13 z2 + 4z = 4 – 6i (z + 2)2 – 22 = 4 – 6i (z + 2)2 = 8 – 6i Let z + 2 = a + bi ⇒ (a + bi)2 = 8 – 6i a2 – b2 + 2abi = 8 – 6i Equating the real and imaginary parts: a2 – b2 = 8 …j

2ab = –6

b = – 3a

…k

Substitute b = – 3a into j:

a2 – 9a2

= 8

a4 – 8a2 – 9 = 0(a2 – 9)(a2 + 1) = 0

a2 = 9 (reject a2 = –1) ⇒ a = ±3, b = ±1 Hence z + 2 = 3 – i or z + 2 = –3 + i ⇒ z = 1 – i or z = –5 + i

When z = 1 – i

|z| = 12 + 12

= 2

arg z = –tan 1112

= – p4 radian

When z = –5 + i|z| = 52 + 12

= 26 arg z = p– tan 11

52= 2.94 radian

14 z2 + z = –9 z2 + z + 9 = 0

z = –1 ± 12 – 4(1)(9)

2(1)

= –1 ± –35

2

[ the roots of the equation are – 12

+ i 35

2

Fully Worked Solution 5


ACE AHEAD Mathematics (T) First Term Updated Edition4

and – 12

– i 35

2 (a) If z is a root ⇒ z* is also a root. Sum of roots, z + z* = –1 [Shown] (b) Product of roots, zz* = 9

|zz*| = |z| |z*| = |z|2 = 9⇒|z| = 3 [Shown]

(c) |z – 1|2 = (z – 1)(z – 1)*= (z – 1)(z* – 1)= zz* – (z + z*) + 1= 9 – (–1) + 1

⇒ |z – 1| = 11 [Shown]

15 z2 + 2z + 4 = 0

z = –2 ± 4 – 16

2 =

–2 ± –32

[ a, b = –1 ± 3 i

Let a = –1 + 3 i = 21– 12

+ 32

i2= 21cos

2p3 + i sin

2p3 2

a3 = 231cos

2p3 + i sin

2p3 2

3

= 8(cos 2p + i sin 2p) [de Moivres’ theorem]= 8(1)= 8

Let b = –1 – 3 i

= 21– 12

– 32

i2

= 21cos 2p3 – i sin

2p3 2

= 23cos 1– 2p3 2 + i sin 1– 2p

3 24b 3 = 233cos 1– 2p

3 2 + i sin 1– 2p3 24

3

= 8[cos (–2p) + i sin (–2p)]= 8(cos 0 + i sin 0)= 8(1 + 0)= 8

[ a3 = b3

16 (a) 1 + i = 2 1cos p4 + i sin

p4 2

(1 + i)18 = 3 2 1cos p4 + i sin

p424

18

= 291cos 9p2 + i sin

9p2 2

= 291cos p2 + i sin

p2 2

[ r = 29 and q = p2

⇒ Modulus = 512

⇒ Argument = p2

(b) (1 + i)(1 – i)

= 2 1cos

p4 + i sin

p4 2

2 1cos p4 – i sin

p4 2

= 1cos

p4 + i sin

p4 2

cos 1– p4 2 + i sin 1– p

4 2

= cos

p4 + i sin

p4

1cos p4 + i sin

p4 2

–1

= 1cos p4 + i sin

p4 2

2

= cos p2 + i sin

p2

(1 + i)9

(1 – i)9 = 1cos

p2 + i sin

p2 2

9

= cos 9p2 + i sin

9p2

= cos p2 + i sin

p2

= i

17 |1 + i| = |1 – i| = 12 + 12

= 2

arg (1 + i) = p4, arg (1 – i) = –

p4

[ 1 + i = 2 1cos p4 + i sin

p42

1 – i = 2 3cos 1– p42 + i sin 1– p

4 24 [ 1 – i = 2 3cos

p4 – i sin

p4 4



) (1 + i)7

(1 + i)9) = |(1 + i)7||(1 – i)9|

= ( 2 )7

( 2 )9

= 12

arg ) (1 + i)7

(1 – i)9 ) = arg (1 + i)7 – arg (1 – i)9

= 71p42 – 91– p42

= 4p= 0

18 (a) 1sin p6

+ i cos p62

6

= 3cos 1p– p62 + i sin 1p–

p624

6

= 1cosp3 + i sin

p32

6

=cos2p + i sin 2p=1

(b) (1 + i)12 = 3 21cos p4 + i sin

p424

12

= ( 2)12(cos 3p + i sin 3p)

= 26(–1)= –64

19 (a) z2 = 2 – 2 3iLet z = a + bi (a + bi)2 = 2 – 2 3ia2 – b2 + 2abi = 2 – 2 3i

Comparing the real parts: a2 – b2 = 2 …j

Comparing the imaginary parts: 2ab = –2 3

b = – 3

a …k

Substituting k into j:

a2 – 1– 3

a 22

= 2

a4 – 2a2 – 3 = 0 (a2 – 3)(a2 + 1) = 0

a2 = 3 or a2 = –1 (reject)a = ± 3

When a = 3, b = – 3 3

= –1

When a = – 3, b = – 1 3 – 32 = 1

⇒ z1 = 3 – i, z2 = – 3 + i

(b) y

xO

( 3, –1)

(– 3, 1)

z2

z1

qa

– 3

3

1

–1

(c) z1 = 3 – i

|z1| = ( 3)2 + (–12) = 2

q = tan–1 1 132 =

p6

∴ Modulus of z = 2

Argument of z = – p6

rad.

z2 = – 3 + i

|z2| = (– 3)2 + 12 = 2

a = tan–1 1 132 =

p6

∴ Modulus of z = 2

Argument of z = p– p6

= 5p6 rad.

20 Let z = x + yi

|z| = 1 ⇒ x2 + y2 = 1x2 + y2 = 1 …j

11 – z

= 11 – (x + yi)

= 11 – x – yi

= 1 – x + yi

(1 – x – yi)(1 – x + yi)

= 1 – x + yi

(1 – x)2 + y2

= 1 – x + yi

1 – 2x + x2 + y2 =

1 – x + yi1 – 2x + 1

(from j)



= 1 – x + yi2(1 – x)

= 1 – x2(1 – x)

+ y

2(1 – x)i = 1

2 +

y2(1 – x)

i

Hence the real part of 11 – z

is 12

.

21 z = cos q + i sin q 1

1 + z2 = 1

1 + (cos q + i sin q)2

= 11 + cos2 q – sin2 q + 2i sin q cos q

= 12 cos2 q + 2i sin q cos q

= 12 cos q (cos q + i sin q)

= 1

2 cos q (cos q + i sin q)–1

= 121

cos q– i sin qcos q 2

= 12(1 – i tan q) (Shown)

11 – z2

= 1

1 – (cos q + i sin )2

= 11 – cos2 q + sin2 q – 2i sin q cos q

= 12 sin2 q – 2i sin q cos q

= 12

3 1sin q(sin q – i cos q) 4

= 12 sin q

1 icos q+ i sin q2

= i(cos q+ i sin q)–1

2 sin q

= i21

cos q– i sin qsin q 2

= 12

(1 + i cot q)

22 (a) z + 1z

= 2 cos q = 1

cos q= 12

⇒ q = p3

z8 + 1z8

= 2 cos 8q

= 2 cos 8p3

= 2 cos 2p3

= 21– 122

=–1 (b) cos 2q= 2 cos2 q– 1

2 cos2 q = cos 2q + 1

= 121z2 + 1

z2 2 + 1

= 123z2 + 1

z2 + 24

= 121z + 1

z22

23 (cos 5q + i sin 5q) = (cos q + i sin q)5

= cos5 q + 5 cos4 q (i sin q) + 10 cos3 q (i sin q)2 + 10 cos2 q (i sin q)3 + 5 cos q (i sin q)4 + (i sin q)5

Comparing the real part:cos 5q = cos5 q + 10 cos3 q (– sin2 q)

+ 5 cos q (sin4 q)= cos5 q – 10 cos3 q (1 – cos2 q)

+ 5 cos q (1 – cos2 q)2

= cos5 q – 10 cos3 q + 10 cos5 q + 5 cos q (1 – 2 cos2 q + cos4 q)

= 11 cos5 q – 10 cos3 q + 5 cos q – 10 cos3 q + 5 cos5 q

= 16 cos5 q – 20 cos3 q + 5 cos q …j

Substitute q = p10 into j:

cos p2 = 16 cos5

p10 – 20 cos3

p10 + 5 cos

p10

0 = 16x5 – 20x3 + 5x where x = cos p10

Since cos p10 ≠ 0,

16x4 – 20x2 + 5 = 0

[ x = cos p10 is a root of 16x4 – 20x2 + 5 = 0



24 Let z1 = z2 = z3 = … = zn = cos q + i sin q z1z2 = (cos q + i sin q)(cos q + i sin q)

= cos2 q – sin2 q + i(2 sin qcos q)= cos 2q + i sin 2q

⇒ (cos q + i sin q)2 = cos 2q + i sin 2qSimilarly z1z2z3

= (cos q + i sin q) (cos q + i sin q) (cos q + i sin q)

= (cos 2q + i sin 2q) (cos q + i sin q)= cos 2q cos q – sin 2q sin q

+ i(sin 2qcos q+ cos 2qsin q)= cos (2q + q) + i sin (2q + q)⇒ (cos q + i sin q)3 = cos 3q + i sin 3q

Hence (cos q + i sin q)n = cos nq + i sin nqfor n = 1, 2, 3, …

cos 5q + i sin 5q = (cos q+ i sin q)5

= cos5 q + 5 cos4 q(i sin q) + 10 cos3 q(i sin q)2 + 10 cos2 q(i sin q)3 + 5 cos q(i sin q)4 + (i sin q)5

Comparing the imaginary parts:sin 5q = 5 cos4 q sin q + 10 cos2 q (–sin q)3

+ (sin q)5

= 5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q

⇒ a = 1, b = –10, c = 5

sin 5qsin q = sin4 q – 10 sin2 q cos2 q+ 5 cos4 q

= (1 – cos2 q)2 – 10(1 – cos2 q)cos2 q + 5 cos4 q

= 1 – 2 cos2 q + cos4 q – 10 cos2 q + 10 cos4 q + 5 cos4 q

= 16 cos4 q – 12 cos2 q + 1

Let x = cos q 16x4 – 12x2 + 1 = 0 becomes16 cos4 q – 12 cos2 q+ 1 = 0

⇒sin 5qsin q = 0

sin 5q = sin kp 5q= p, 2p, 3p, 4p

q= 15p, 2

5p, 3

5p, 4

5p

Hence the solutions are

x = cos p5

, cos 2p5

, cos 3p5 , cos

4p5

16 cos4 q – 12 cos2 q + 1 = 0

cos2 q = 12 ± 144 – 4(16)(1)

2(16)

= 3 ± 5 8

y

xO

52p

5p

cos2 p5

+ cos2 2p5

= 3 + 5

8 +

3 + 5 8

= 68

= 34

25 (a) w3 = 1 (w3)2 = 1 ⇒ (w2)3 = 1

⇒ w2 = 1 [Shown] (b) w3 – 1 = 0 ⇒ (w – 1)(w2 + w + 1) = 0

⇒ w = – 12

± 3

2i

⇒ w2 = 1– 12

+ 3

2i2

2

= – 12

– 3

2i

1 + w + w2 = 1 + 1– 12

+ 3

2i2

+ 1– 12

– 3

2i2

= 0 [Shown

(c) w5 + w7 = (w3)(w2) + (w6)(w) = w2 + w = –1


ACE AHEAD Mathematics (T) First Term Updated Edition8

26 w4 = –16i

= 161cos 3p2

+ i sin 3p2 2

= 163cos 13p2 + 2kp2 + i sin 13p2

+ 2kp24⇒ w = 23cos 13p

2 + 2kp2 + i sin 13p

2 + 2kp24

14

= 23cos 13p8

+ kp2 2 + i sin 13p

8 +

kp2 24

(De Moivres’ theorem)

When k = 0, w1 = 21cos 3p8

+ i sin 3p8 2

When k = 1, w2 = 23cos 13p8 + p22

+ i sin 13p8 + p224

= 21cos 7p8

+ i sin 7p8 2

When k = 2, w3 = 23cos 13p8 + p2 + i sin 13p8 + p24 = 21cos

11p8

+ i sin 11p

8 2

When k = 3, w4 = 23cos 13p8

+ 3p2 2

+ i sin 13p8

+ 3p2 24

= 2 1cos 15p

8 + i sin

15p8 2

Hence, the roots of w4 = –16i are

21cos 3p8

+ i sin 3p8 2, 21cos

7p8

+ i sin 3p8 2,

21cos 11p

8 + i sin 11p

8 2 and

21cos 15p

8 + i sin

15p8 2.

Imaginary

Real

2

–2

–2 2O

83p8

7p

811p

815p

w1

w4

w3

w2

Chapter 4

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