chap 4 complex numbers focus exam ace

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

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, z

2* = 1 – i

1z*

1

+ z1z

2* = 1

3 + 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, z

2* = 2 – i

z1z

2* + z

1*z

2 = (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

= i2i + 2

+ i2i – 2

= i2i + 2

– i2 – 2i

= i1 1z*

– 1z2 [Shown]

5 (a) z3 = z

1z

2

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

|z3| = 112 + (–2)2

= 11.18 (b) arg z

3 = tan–1 1–2

112= –tan–1 1 2

112= –0.18 radian

(c) Imaginary

RealO

(11, –2)z3

6 3 – ai

1 – 3 i =

3 – ai

1 – 3 i

1 + 3 i1 + 3 i

= 3 + a 3 + (3 – a)i

1 + 3

= 3 (a + 1) + (3 – a)i

4

Chap-4.indd 1 3/1/2012 10:56:33 AM


ACE AHEAD Mathematics (T) First Term2

3 – ai

1 – 3 i is a real number ⇒ 3 – a = 0

a = 3

Hence the real number is 3 (3 + 1)

4 = 3

7 z1z

2 = (1 – 3i)( 3 + i)

= 3 + 3 + (1 – 3)i

= 2 3 – 2i

r = |z1z

2| = (2 3)2 + (–2)2

= 12 + 4= 4

q = –tan–1 1 22 3 2

= – p6

[ z1z

2 = 43cos 1– p

62 + i sin 1– p624

Imaginary

RealO

r

(2 3, –2)

q

8 (x + iy)2 = 4i x2 – y2 + 2xyi = 4i Comparing the real and imaginary parts: x2 – y2 = 0 …(1) 2xy = 4

y = 2x …(2)

Substitute y = 2x into (1):

x2 – 12x2

2

= 0

x4 – 4 = 0 (x2 + 2)(x2 – 2) = 0 x2 = –2 or x2 = 2 [ x = ± 2 (reject x2 = –2)

Substitute x = ± 2 into (2):

y = ± 22

= ± 2 Hence x = ± 2, y = ± 2

9 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 …(1)

⇒ 5a = 20a = 4

Substitute a = 4 into (1):16 + b2 + 5b = 10b2 + 5b + 6 = 0

(b + 2)(b + 3) = 0b = –2 or b = –3

[ z = 4 – 2i or z = 4 – 3i

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

a2 – b2 = 1 …(1)2ab = 2 2

b = 2

a …(2)

Substitute b = 2

a into (1):

a2 – 2a2

= 1

a4 – a2 – 2 = 0(a2 – 2)(a2 + 1) = 0

a2 = 2 (reject a2 = –1)a = ± 2

Substitute a = ± 2 into (2) ⇒ b = ± 2 2

= 1 Hence z

1 = 2 – i and z

2 = – 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 1

2 2 = –0.615 radian

arg z2 = p – tan–1 1 1

2 2 = 2.526 radian

Chap-4.indd 2 3/1/2012 10:56:33 AM


Fully Worked Solution 3

11 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 …(1)

2ab = –6

b = – 3a

…(2)

Substitute b = – 3a into (1):

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

12 z2 + z = –9 z2 + z + 9 = 0

z = –1 ± 12 – 4(1)(9)

2(1)

= –1 ± –35

2 [ the roots of the equation are – 1

2 +

i 352

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]

y

xO–1 1 2 4–3

Locus of (c)

Locus of (a)

Locus of (b)

21x = −

–2 3

13 (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 2

i22

= – 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

14 (a) |z + 1 – i|2 = 2 |z – (–1 + i)| = 2 The locus of z is a circle with centre

(–1, 1) and radius 2.

Imaginary

RealO–2

22

Chap-4.indd 3 3/1/2012 10:56:34 AM


(b) |z + 1| = |z – i| The locus of z is the perpendicular

bisector of line joining (–1, 0) and (0, 1).

Imaginary

RealO–1

1

(c) z + z* + 2 = 0 x + iy + x – iy + 2 = 0

2x + 2 = 0x = –1

Imaginary

RealO

x = –1

–1

15 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)i

When x = y, z + iz = (y – y) + (y + y)i= 2yi

⇒ R lies on the imaginary axis.

16 (a) |z + 3| = 3|z – (–3)| = 3The locus of P is a circle with radius (–3, 0) and radius 3.

Imaginary

Real

3

–6 –3

–3

O

(b) – 12

p < arg 11z 2 < p

4

⇒ – p2

< –arg z < p4

– p4

< arg z < p2

4p

Imaginary

ReRealO

(c) –p < arg z < – p2

⇒ p < arg z < p2

Imaginary

RealO


Chap-4.indd 4 3/1/2012 10:56:34 AM



(d) |z – i| > |z + 1|If z = –1, |z – i| = |–1 – i|

= 12 + 12

= 2|z + 1| = |–1 + 1|

= 0z = –1 satisfies the inequality.⇒ (–1, 0) lies in the region required.

Imaginary

RealO(–1, 0)

(0, 1)

17 (a) The locus of z is the perpendicular bisector of the line segment joining the points (2, 0) and (0, 4).

Imaginary

RealO 2

4 locus

(b) Let the points A and B be A(–1, 0) and B(2, 0).The locus required is the locus of point P where BP = 2AP.From diagram, BL = 2AL and BK = 2AK.[ L (0, 0) K (– 4, 0)[ The locus of z is the circle with

centre (–2, 0) and radius 2. ⇒ |z + 2| = 2

Imaginary

RealO−4 −1

P

K

2

L BA

locus

18 z2 + 2z + 4 = 0

z = –2 ± 4 – 162

= –2 ± –32

[ a, b = –1 ± 3 i

Let a = –1 + 3 i = 21– 12

+ 32

i2= 21cos 2p

3 + i sin 2p

3 2a3 = 231cos 2p

3 + i sin 2p

3 23

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

Let b = –1 – 3 i

= 21– 12

– 32

i2 = 21cos 2p

3 – i sin 2p

3 2 = 23cos 1– 2p

3 2 + i sin 1– 2p3 24

b 3 = 233cos 1– 2p3 2 + i sin 1– 2p

3 243

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

[ a3 = b3

19 (a) 1 + i = 2 1cos p4

+ i sin p42

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

+ i sin p424

18

= 291cos 9p2

+ i sin 9p2 2

= 291cos p2

+ i sin p22

[ r = 29 and q = p2

⇒ Modulus = 512

⇒ Argument = p2

(b) (1 + i)(1 – i)

= 2 1cos p

4 + i sin p

422 1cos p

4 – i sin p

42

Chap-4.indd 5 3/1/2012 10:56:36 AM




= 1cos p

4 + i sin p

42cos 1– p

42 + i sin 1– p42

= cos p

4 + i sin p

4

1cos p4

+ i sin p42

–1

= 1cos p4

+ i sin p42

2

= cos p2

+ i sin p2

(1 + i)9

(1 – i)9 = 1cos p

2 + i sin p

229

= cos 9p2

+ i sin 9p2

= cos p2

+ i sin p2

= i

20 |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

424 [ 1 – i = 2 3cos p

4 – i sin p

44

)(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

21 (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 …(1)

Substitute q = p10

into (1):

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

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

Chap-4.indd 6 3/1/2012 10:56:37 AM



= 123z2 + 1

z2 + 24

= 121z + 1

z22

23 (a) 1sin p6

+ i cos p62

6

= 3cos 1p – p62 + i sin 1p – p

6246

= 1cos p3

+ i sin p32

6

= cos 2p + 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

24 (a) Imaginary

RealO–3

–3

3

3(3, 4)

P

|z – (3 + 4i)| is minimum when the point P is nearest to the point (3, 4).⇒ Minimum value of

|z – (3 + 4i)| = 5 – 3= 2

(b) Imaginary

RealO

–12

5

4

4

|z – (5 – 12i)| < 4 is the locus of the point P which represents the complex

number z that lies on or in the circle with centre (5, –12) and radius 4 units.⇒ 13 – 4 < z < 13 + 4

9 < z < 17

25 (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 …(1)

Comparing the imaginary parts: 2ab = –2 3

b = – 3 a

…(2)

Substituting (2) into (1):

a2 – 1– 3 a 2

2

= 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, z

2 = – 3 + i

(b) y

xO

( 3, –1)

(– 3, 1)

z2

z1

qa

– 3

3

1

–1

(c) z1 = 3 – i

|z1| = ( 3)2 + (–1)2 = 2

q = tan–1 1 1

32 = p6

∴ Modulus of z = 2

Argument of z = – p6

rad.

Chap-4.indd 7 3/1/2012 10:56:38 AM



z2 = – 3 + i

|z2| = (– 3)2 + 12 = 2

a = tan–1 1 1

32 = p6

∴ Modulus of z = 2

Argument of z = p – p6

= 5p6

rad.

26 Let z = x + yi

|z| = 1 ⇒ x2 + y2 = 1x2 + y2 = 1 …(1)

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 (1))

= 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

.

27 |z – 2| + |z + 2| = 5

(x – 2)2 + y2 = 5 – (x + 2)2 + y2

(x – 2)2 + y2 = 25 + (x + 2)2 + y2 – 10 (x + 2)2 + y2

x2 – 4x + 4 + y2 = 25 + x2 + 4x + 4

+ y2 – 10 (x + 2)2 + y2

10 (x + 2)2 + y2 = 8x + 25100(x2 + 4x + 4 + y2) = 64x2 + 400x + 625100x2 + 400x + 400 + 100y2 = 64x2 + 400x + 625

36x2 + 100y2 – 225 = 0(6x)2 + (10y)2 = (15)2

16x152

2

+ 110y15 2

2

= 1

⇒ The locus of P is an ellipse with centre

(0, 0) and vertices ± 152, 02.

@ @ @ @ @

@

@@

y

x

23

25

2–5

2–3

O

|z| = )z – 310

+ 310

i)x2 + y2 = 1x – 3

1022

+ 1y + 3102

2

x2 + y2 = x2 – 35

x + 9100

+ y2 + 35

y + 9100

35

x – 35

y = 18100

= 950

30x – 30y = 910x – 10y = 3

10y = 10x – 3100y2 = (10x – 3)2 …(1)

36x2 + 100y2 = 225 …(2)(2) – (1): 36x2 = 225 – (100x2 – 60x + 9)

36x2 = 225 – 100x2 + 60x – 9136x2 – 60x – 216 = 0

34x2 – 15x – 54 = 0 (17x + 18)(2x – 3) = 0

x = – 1817

, x = 32

When x = – 1817

, 101– 18172 – 10y = 3

⇒ 10y = – 18017

– 3 = – 23117

y = – 231170

When x = 32

, 101322 – 10y = 3

⇒ 15 – 3 = 10y

y = 1210

= 65

The points on the locus are 1– 1817

, – 2311702

and 132, 652.

28 Let z1 = z

2 = z

3 = … = z

n = cos q + i sin q

z1z

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

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

Chap-4.indd 8 3/1/2012 10:56:40 AM



Similarly z1z

2z

3

= (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 2q cos q + cos 2q sin 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 nq for 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 = 15

p, 25

p, 35

p, 45

p

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

29 (a) y + ixx + iy

= x + iy

y + ix = x2 – y2 + 2xyi⇒ 2xy = x

y = 12

⇒ x2 + y2 = y

x2 – 14

= 12

x2 = 34

x = 3 2

\ x = 3

2, y = 1

2

(b) z2 =

3 2

+ 12

i ⇒ z2 = cos p

6 + i sin p

6

z3 = 1

2 + 3

2i ⇒ z

3 = cos p

3 + i sin p

3

(c) z1 + z

2 + … + z

n

= 1 + 1cos p6

+ i sin p62

+ 1cos p6

+ i sin p62

2

+ …

+ 1cos p6

+ i sin p62

n

Chap-4.indd 9 3/1/2012 10:56:42 AM



= 131cos p

6 + i sin p

62n

– 141cos p

6 + i sin p

6 – 12

= cos np

6 + i sin np

6 – 1

cos p6

+ i sin p6

– 1

Therefore cos np6

+ i sin np6

– 1 = 0

⇒ cos np6

– 1 = 0 and sin np6

= 0

cos np6

= 1

= cos 2kp where k = 1, 2, 3, …n6

= 2k

n = 12, 24, 36, 48, 60, …

sin np6

= 0

= sin kpn6

= k

n = 6, 12, 18, 24, …Hence the smallest positive integer of n is 12.

(d) z1z

2z

3 … z

12

= (1)1cos p6

+ i sin p62

1

1cos p6

+ i sin p62

2

…

1cos p6

+ i sin p62

12

= 1cos p6

+ i sin p62

1 + 2 + 3 + … + 12

= 1cos p6

+ i sin p62

78

= cos 13p + i sin 13p= cos p + i sin p= –1

30 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)

= 12 cos q

(cos q + i sin q)–1

= 121cos q – i sin q

cos q 2=

12(1 – i tan q) (Shown)

11 – z2

= 11 – (cos q + i sin q)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

= i21cos q – i sin q

sin q 2=

12

(1 + i cot q)

31 (z – ia)3 = i3

= 1cos p2

+ i sin p22

3

= cos 3p2

+ i sin 3p2

= cos 13p2

+ 2pk2 + i sin 13p2

+ 2pk2z – ia = 3cos 13p

2 + 2pk2 + i sin 13p

2 + 2pk24

13

= cos 1p2 + 2pk3 2 + i sin 1p2 + 2pk

3 2where k = 0, 1, 2

When k = 0, z1 – ia = i

⇒ z1 = i(a + 1)

When k = 1,

z2 – ia = cos 1p2 + 2p

3 2 + i sin 1p2 + 2p3 2

= cos 7p6

+ i sin 7p6

= – 3

2 –

12

i

⇒ z2 = –

3 2

+ 1a – 122i

Chap-4.indd 10 3/1/2012 10:56:43 AM



When k = 2,

z3 – ia = cos 1p2 + 4p

3 2 + i sin 1p2 + 4p3 2

= cos 11p6

+ i sin 11p6

= 3

2 – 1

2i

⇒ z3 =

3 2

+ 1a – 122i

[ The roots of (z – ia)3 = i3 are i(a + 1),

– 3

2 + 1a –

122i and

3 2

+ 1a – 122i

(a)

A (a + 1)

O

Imaginary

Real

32

, a – 12B

32

, a – 12C–

BC = 3

AB = BC = 1 3 2 2

2

+ 1a + 1 – a + 122

2

= 34

+ 94

= 3

Hence A, B and C which represent the roots of the equation form an equilateral triangle.

(b) [z – (1 + i)]3 = (2i)3

= 23i3

⇒ z1 – (1 + i) = 2i,

z2 – (1 + i) = 21–

3 2

– 12

i2,z

3 – (1 + i) = 21 3

2 –

12

i2⇒ z

1 = 2i + 1 + i, z

2 = – 3 – i + 1 + i,

= 1 + 3i = – 3 + 1 z

3 = 3 – i + 1 – i= 3 + 1

Hence the solutions are 1 + 3i, – 3 + 1 and 3 + 1

(c) ax2 = bx + c = 0

x = –b ± b2 – 4ac

2a

= –b ± ( b2 – 4ac)i

2a where b2 < 4ac

Hence, if w = –b2a

+ ( b2 – 4ac)i

2a is a

root, then its conjugate

w* = – b2a

– ( b2 – 4ac)i

2a is also a root.

The roots of (z – ia)3 = i3 are i(a + 1),

– 32

+ 1a – 122 i and 3

2 + 1a – 1

22 iLet a = 1∴ The roots of (z – i)3 = i3 are 2i,

– 32

+ 12

i and 32

+ 12

i

Let z1 = 2i ⇒ z

1* = –2i

Sum of roots = z1 + z

1* = 0

Product of roots = z1z

1* = 4

∴ x2 + 4 = 0

Let z2 =

– 32

+ 12

i ⇒ z2* =

– 32

– 12

i


2* = – 3


2*

1– 32

+ 12

i2 = 1– 3 2

– 12 i2 = 1

∴ x2 + 3 x + 1 = 0

Let z3 = 3

2 + 1

2 i ⇒ z

3* =

32

– 12

i


3* = 3


3*

= 1 32

+ 12

i2 1 32

– 12

i2 = 1

∴ x2 – 3 x + 1 = 0⇒ (x2 + 4)(x2 + 3 x + 1)(x2 – 3 x + 1) = 0(x2 + 4)[(x2 + 1) + 3 x][(x2 + 1)

– 3 x] = 0(x2 + 4)[(x2 + 1)2 – 3x2] = 0x2(x2 + 1)2 – 3x4 + 4(x2 + 1)2 – 12x2 = 0x2 (x4 + 2x2 + 1) – 3x4 + 4(x4 + 2x2 + 1) – 12x2 = 0x6 + 2x4 + x2 – 3x4 + 4x4 + 8x2 + 4 – 12x2 = 0x6 + 3x4 – 3x2 + 4 = 0

Chap-4.indd 11 3/1/2012 10:56:46 AM



32 1 < |z + 2 – 2i| < 3|z – (–2 + 2i)| = 3 is a circle with centre (–2, 2) and radius 3 unit.|z – (–2 + 2i)| = 1 is a circle with centre (–2, 2) and radius 1 unit.

Imaginary

Real

1

2

3

4

5

–4–5 –3 –2 –1 O

|z + 2 – 2i | = 1

|z + 2 – 2i | = 3

P

C

OC = 22 + 22 = 2 2 CP = radius = 3OP = 3 + 2 2

∴ Range of |z| is 0 < |z| < 3 + 2 2

33 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

34 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 + kp

2 24(De Moivres’ theorem)

When k = 0, w1 = 21cos 3p

8 + i sin 3p

8 2When k = 1, w

2 = 23cos 13p

8 + p

22 + i sin 13p

8 + p

224= 21cos 7p

8 + i sin 7p

8 2When k = 2, w

3 = 23cos 13p

8 + p2

+ i sin 13p8

+ p24 = 21cos 11p

8 + i sin 11p

8 2When k = 3, w

4 = 23cos 13p

8 + 3p

2 2 + i sin 13p

8 + 3p

2 24 = 2 1cos 15p

8 + i sin 15p

8 2Hence, the roots of w4 = –16i are

21cos 3p8

+ i sin 3p8 2, 21cos 7p

8 + i sin 3p

8 2, 21cos 11p

8 + i sin 11p

8 2 and

21cos 15p8

+ i sin 15p8 2.

Imaginary

Real

2

–2

–2 2O

83p8

7p

811p

815p

w1

w4

w3

w2

Chap-4.indd 12 3/1/2012 10:56:47 AM

chap 4 complex numbers focus exam ace

Education

z2 z

root z

iz z

shown c z

a2 b2

substitute b

locus of b

z2 z1 c z1