© 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
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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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
(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
ACE AHEAD Mathematics (T) First Term4
Chap-4.indd 4 3/1/2012 10:56:34 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 5
(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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) First Term6
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
= 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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 7
= 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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) First Term8
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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 9
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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) First Term10
= 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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 11
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
Sum of roots = z2 + z
2* = – 3
Product of roots = z2z
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
Sum of roots = z3 + z
3* = 3
Product of roots = z3z
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
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) First Term12
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