Unit 2 Chapter 1 Answers PDF

 

Page 1 of 58 

Chapter 1 Complex numbers Try these 1.1

(a) (i) Re (5 + 4i) = 5 Im(5 + 4i) = 4 (ii) Re(4 + 7i) = 4 Im(4 + 7i) = 7 (iii) Re(5 x + i (3 xy) = 5 x Im(5 x + i (3 xy) = 3 xy (iv) Re(7 x2 + y + i (3 x – 2 y)) = 7 x2 + y  Im(7 x2 + y + i (3 x – 2 y)) = 3 x – 2 y  (v) 7i2 – 4i = –7 –4i since i2 = –1 Re(7i2 – 4i) = –7 Im(7i2 – 4i) = –4 (b) (i) 4 xi + 3 yi – 2 x  = –2 x + i (4 x + 3 y) Real part = –2 x  Imaginary part = 4 x + 3 y  (ii) (cosθ  )i + sinθ   θ 

Real part =part sin =  cosθ   Imaginary (iii) 4 sin θ  –  – (3 cosθ )i  Re(4 sinθ  –  – (3 cosθ )i) = 4 sinθ   Im(4 sinθ  –  – (3 cosθ i)) = –3 cosθ   2 (iv) 8 cos θ  +  + 7cos θ   + + i sin3 θ   – – i sin4 θ   Real part = 8 cos2θ  +7  +7 cosθ   Imaginary part = sin3θ  –  – sin4 θ   (v) 8 cos2θ  i2 + 7 sin3θ  i3 + 4i4 cos2θ  +  + 7 sinθ   2 3 = – 8 cos θ   – – i7 sin θ  +  + 4 cos2θ  +  + 7 sinθ   2 = – 8 cos θ  +  + 4 cos 2θ  +  + 7 sinθ   – – i 7 sin3 θ   Real part = – 8 cos 2θ  +  + 4 cos2θ  +  + 7 sinθ   3 Imaginary part = –7 sin  θ  

Try this 1.2

Let 3 − 4i = x + iiyy   2 2 ∴ 3 – 4i = x  – y  + i (2 xy) Equating real and imaginary parts: 2 2  x  – y   =3 2 xy  = 4 From [2]  y =

[1] [2]

4 2 =   2 x x

Substituting for y in [1]  x 2 −  x

4

= 3 

 x 2

4

− 4 = 3x 2  

4

2

 x 2− 3 x −24 = 0   ( x − 4)( x + 1) = 0  

Unit 2 Answers: Chapter 1

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Page 2 of 58 

 x = 4 since x, y  x = ± 2   2

∈  

2 = 1  2 2 When  x = −2, y = = 1  −2 ∴ 3 + 4i = 2 + i, − 2 − i   When  x = 2, y =

Exercise 1A 1 

 z1 = 2 + 4i, z2 = 3 + 5i  (a)  z1 + z2 = 2 + 4i + 3 + 5 i = (2 + 3) + (4i + 5i) = 5 + 9i  (b)  z1 – z2 = (2 + 4i) – (3 + 5i) = (2 – 3) + (4i – 5i) = –1 – i  (c)  z1  zz2 = (2 + 4i) (3 + 5i) = 6 + 10i + 12i + 20i2 = 6 – 20 + 22i 

= –14 + 22i   z 2  + 4 i (d) 1 =    z 2 3 + 5i 2 + 4 i 3 − 5i = ×   3 + 5i 3 − 5i

2 

3 

6 − 10 10 i + 12i − 20i 2 =   9 + 25  6 + 20 + 2i   = 34 26 2 = + i  34 34 13 1 = + i  17 z  =1(3 7 + i) – (4 – 3i) (a)  z1 – 2 = 3 – 4 + i + 3i  = –1 + 4i  (b)  z1 + z3 – z4 = 3 + i + (–1 + 2i) – (–2 – 5i) = 3 – 1 + 2 + i + 2i + 5i  = 4 + 8i  (c)  z1* z2 = (3 – i) (4 – 3i) = 12 – 9i – 4i + 3i2 = 12 – 3 – 9 i – 4i  = 9 – 13i  (a)  z1 + z2 = 3 + i + 4 – 3i  = 7 – 2i  (b)  z3  zz4 = (–1 + 2i) (–2 – 5i) 2

= 22 + + 10 5i –+4ii  – 10i =

Unit 2 Answers: Chapter 1

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= 12 + i   z3* −  1 − 2i (c)   =  z4* − 2 + 5i − 1 − 2 i − 2 − 5i = ×   − 2 + 5i − 2 − 5i 2 + 5i + 4i + 10i 2   4 + 25 =

=

4 

−  8 + 9i

  29 −8 9 = + i  29 29 (a)  z1  zz2  zz3 = (3 + i) (4 – 3i) (–1 + 2i)  = (12 – 9i + 4i – 3i2) (–1 + 2i) = (15 – 5i) (–1 + 2i) = –15 + 30i + 5i – 10i2  = –5 + 35i  (b)  z2  zz3 + z1  zz4  = (4 – 3i) (–1 + 2i) + (3 + i) (–2 – 5i)  = –4 + 8i + 3i – 6i2 – 6 – 15i – 2i – 5i2 = –4 + 6 – 6 + 5 + 8 i + 3i – 15i – 2i  i

= z *1+–z6*       3 − i + 4 + 3i (c) 1 * * 2 =    z 3 z 4 (–1 – 2i ) (−2 + 5i ) 7 + 2i   2 − 5i + 4i − 10  i 2 7 + 2i   = 12 − i 7 + 2i 12 + i = ×   12 − i 12 + i =

84 + 7i + 24 i + 2 i 2 =   144 + 1  82 + 31i   =

5 

82145 31 i  = + 145 145  z 1   3 + i (a)   =  z 2 4 − 3i 3 + i 4 + 3i = ×   4 − 3i 4 + 3i 12 + 9i + 4i + 3i 2   = 16 + 9 9 + 13i 9 13 = = + i  25 25 25 25  z + z 2   (b) 1  z z 3

4

Unit 2 Answers: Chapter 1

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3 + i + 4 − 3i   (−1 + 2i ) (−2 − 5i ) 7 − 2i   = 2 + 5i − 4i − 10  i 2 7 − 2i =   12 + i 7 − 2i 12 − i = ×   12 + i 12 − i =

84 − 7i − 24 i + 2 i 2   = 144 + 1 82 31 i  = − 145 145  z 1 z 2 (4 − 3i )  (3 + i ) (4 (c) =   3 + i + 4 − 3i  z 1 + z 2 12 − 9 i + 4 i − 3i 2   = 7 − 2i 15 − 5i 7 + 2i = ×   7 − 2i 7 + 2i

(d)

6 

(a) (b) (c) (d) (e)

105 + 30i − 35i − 10 i 2   = 49 + 4  115 − 5i   = 53 115 5i = −   53 53  z 1 + z 2     3 + i + 4 − 3i   =  z 3 + z 4 − 1 + 2i − 2 − 5i 7 − 2i =   − 3 − 3i − 3 + 3i 7 − 2i = ×   − 3 − 3i − 3 + 3i −21 + 21 21i + 6i − 6i 2 = 9+9 −15 27   = + i 18 18 −5 3 = + i 6 2 2 6 12 i  = (i )  = (–1)6 = 1 15 2 7 7 i  = i × (i )  = i (–1)  = – i  i21 = i × (i2)10 = i (–1)10 = i  4 4 4 = 2 4 = = 4  8 i (i ) ( − 1)4 5 5 5 = = =5  20 2 10 10 i (i ) (−1)

Unit 2 Answers: Chapter 1

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Page 5 of 58 

7

1 + 3i   1 − 2i 1 + 3i 1 + 2 i   = × 1 − 2i 1 + 2 i

(a)  z =

1 + 2i + 3i + 6 i 2   12 + 22 −5 + 5i = = −1 + i   5  z2 = (–1 + i)2  = (–1 + i) (–1 + i) = 1 – i – i + i2 = –2i  1 1 (b)  z  – = − 1 + i −   −1+ i  z (−1 − i )   = −1 + i − (−1 − i)(−1 + i)

=

   −1− i  = − 1+ i −   2 2    1 +1  = − 1+ i + 1 + 1 i   2

2

1 3 i  2 2 2 8   z  = – 5 + 12 i  ⇒  zz = − 5 + 12i  

=− +

 Now

− 5 + 12i  = a + bi 

⇒ –5 + 12i = (a + bi)2 ⇒ –5 + 12i = a2 + 2abi + b2i2 ⇒ –5 + 12i = a2 – b2 + 2abi  Equating real and imaginary parts: 2 2 a  – b  = –5 [1] 2ab = 12 [2] From [2]: b = 12 = 6   2a a Substitute into [1] 2

  6  ⇒ a −  = −5  a 36 ⇒ a2 − 2 = − 5   2

a

⇒ a − 36 = − 5a 2   ⇒ a4 + 5a2 – 36 = 0 ⇒ (a2 + 9) (a2 – 4) = 0 ∴ a2 + 9 = 0, a2 – 4 = 0 Since a is real, a = ±2 4

6 When a = 2, b = 2 = 3  

Unit 2 Answers: Chapter 1

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Page 6 of 58 

6 = −3  −2 ∴  zz = 2 + 3 i,  zz = –2 – 3i  1 1 1 = +  

When a = –2, b =

9 

u

1

v

=

w

1

+

1

 

1 − 2i 3 + i 1 + 2i 3−i = 2 2+ 2 2  1 +2 3 +1 1 2 3 1 = + i+ − i  5 5 10 10 10 1 3 = + i  2 10 1 1 3 Since = + i  u 2 10 1   ⇒ u = 1 3 + i 2 10 1 3 2 − 10 i   = 2 2  1  3    +   2 10 1 3 − i 2 10   = 1 9 + 4 100 1 3 − i 2 10   = 34 100 100  1 3  =  − i   34  2 10  50 30 = − i  34 34 25 15 = − i  17 17 2 − i 6 + 8i   − 10   z = 1+ i x + i 2 − i 1 − i 6 + 8i x − i   = × − × 1+ i 1− i x + i x − i 2 − 2 i − i + i 2 6 x − 6 i + 8x i − 8i 2   = − 1+1  x 2 + 12   1 − 3i 6x + 8 + i (8x − 6)   = −  x 2 + 1 2 u

Unit 2 Answers: Chapter 1

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 1 6 x + 8   −3 8 x − 6  = − 2 +i − 2    2  x + 1   2 x + 1  1 6 x + 8   − 2  x 2 + 1 −3 8 x − 6 − 2   Im (  zz) = + 2z) =  x Since Re (  z Im (  zz1) 1 6 x + 8 − 3 8 x − 6   ⇒ − 2 = − 2 2  x + 1 2 x +1 6 x + 8 6 − 8 x   ⇒2− 2 = 2  x + 1 x +1 ⇒ 2 x2 + 2 – 6 x – 8 + 8 x – 6 = 0 ⇒ 2 x2 + 2 x – 12 = 0  x2 + x – 6 = 0 (  xx + 3) (  xx – 2) = 0  x = –3, 2 bii   11  (a) 3 + 4 i = a + b 3 + 4i = (a + bi)2 Re (  zz) =

2

2 2

bi 2 33 + + 44ii =  = a a  +  – b22abi  +  + 2abi   Equating real and imaginary parts: [1] ⇒ a2 – b2 = 3 2ab = 4 4 2 From [2] ⇒ b = =   2a a Substitute into [1] 2  2 2 a  –    = 3  a 4 a2 – 2  = 3

[2]

a a  – 3a2 – 4 = 0 (a2 – 4) (a2 + 1) = 0 a2 = 4, a2 = –1 4

Since a is real, a = ±2 2 When a = 2, b = = 1   2 2 When a = –2, b = = −1  −2 ∴  3 + 4i = 2 + i, − 2 − i   (b)

24 − 10i = a + bbii   24 – 10i = (a + bi)2 24 – 10i = a2 – b2 + 2abi ∴ a2 – b2 = 24 2ab = –10 −5 b From [2] ⇒   = a  

Unit 2 Answers: Chapter 1

[1] [2]

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Substitute into [1] 2  −5  2 a  –    = 24



2

a  –

a

25



= 24  

a2 4 2 a  – 24a  – 25 = 0

(a22 – 25) (2a2 + 1) = 0 a  = 25, a  = –1 Since a is real, a = ±5 −5 When a = 5, b = = − 1  5 −5 When a = –5, b = = 1  −5 ∴  24 − 10 10i = 5 − i , –5 + i

Exercise 1B 1 

2

 z  + 16 = 0

 z2 = –16

 z = ±   − 16  = 4i, – 4i 

2 

 z2  – 8 z + 17 = 0  z =

 8 ± 64 − (4) (4) (17) (1)   2(1)

8 ± −4   2 8 ± 2i = =4±i  2 ∴  zz = 4 + i or 4 – i  3   z2 – 4 z + 5 = 0 4 ± 16 − 20 20  z =   2 4 4 = ± −   2 4 ± 2i   = 2 = 2 ± i  ∴  zz = 2 + i or 2 – i  4   z2 – 6 z + 13 = 0  6 + 36 − (4) (13) (1)    z = 2(1)

=

= =

6 ± −16 2   6 ± 4i

= 3 ±22i   Unit 2 Answers: Chapter 1

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Page 9 of 58 

5 

 z = 3 + 2 i or 3 – 2i  2  z  – 10 z + 31 = 0.  z =

 10 ± 100 − 4(31) (1)   2(1)

10 ± −24   2 10 ± 2 6i = = 5 ± 6i   2 ∴  zz = 5 + 6i  or 5 – 6 i   2  z  + 1 = ( z + i) (  zz – i) 2  z  – 2 z + 2 = ( z – 1 – i) (  zz – 1 + i)  z2 – 6 z + 25 = ( z – (3 + 4i)) (  zz – (3 – 4i)) = (  zz – 3 – 4i) (  zz – 3 + 4i) 4 2 2  z  – z  – 2 z + 2 = ( z – 1)  (  zz + 1 + i) (z + 1 − i). Let 2 i  = a + bi  ⇒ 2i = (a + bi)2 2i = a2 + 2abi + b2i2 2i = a2 – b2 + 2abi  Equating real and imaginary parts

=

6  7  8  9  10 

2

2

a b 2 = 0  – = 2ab

From [2] b =

[1]

[2]

2 1 =   2a a

2

 1 a  –   = 0  a 2

2

a  – 4

1 a2

 = 0

a  – 1 = 0 (a2 – 1) (a2 + 1) = 0 2 2 a  – 1 = 0, a  + 1 = 0 Since a is real, a = ± 1

When a = 1, b =

1

= 1 

11 When a = –1, b =  = –1 −1 ∴ 2i = 1 + i  or –1 – i   z2 – (3 + 5 i)  zz – 4 + 7i = 0 3 + 5i ± (3 + 5i )2 − 4 ( −4 + 7i ) 3 + 5i ± −16 + 30i + 16 − 28    z = = 2 2 3 + 5i ± 2 i 3 + 5i + 1 + i 3 + 5i − 1 − i   = ∴ z = , 2 2 2 = 2 + 3i or 1 + 2i  2 u  = –60 – 32i  11  ⇒ (  xx + iy)2 = –60 + 32i  ⇒  xx2 – y2 + 2 xyi = –60 + 32i  Equating real and imaginary parts

 x2 – y2 = – 60

Unit 2 Answers: Chapter 1

[1]

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Page 10 of 58 

2 xy = + 32

[2]

32 16 =   2 x x Substituting into [1]

From [2]  y = 2

 16   x  –   = – 60   x  2

 x2 – 256  = – 60  x 2  x4 + 60 x2 – 256 = 0 (  xx2 – 4) ( x2 + 64) = 0 2 2  x  = 4, x  = –64 Since x ∈ ℝ, x = ±2

When x = 2,  y =

16 =8  2

16 = −8  −2 ∴ u = 2 + 8i or –2 – 8i  2  z  – (3 – 2i) z + 5 – 5i = 0  x = –2,  y

 z =

=

3 − 2i ± (3 − 2i )2 − 4(5 − 5i )

 

2 3 − 2 i ± 9 − 12 i + 4 i 2 − 20 + 20 i   = 2  3 − 2 i ± −15 + 8i   = 2 Since −60 + 32i = ± (2 + 8i)   4 (−15 + 8i ) = ± (2 + 8i )   ∴ −15 + 8i = ± (1 + 4i )  

  3 − 2i + 1 + 4i 3 − 2i − 1 − 4i ∴   z =   or  2

2

4 + 2i 2 − 6i   or  2 2 = 2 + i or 1 – 3i  3 2 3z  – 23 z  + 52 z + 20 = 0 12  2  f (  zz) = 3 z3 –23 z  + 52 z + 20  f (4 (4 + 2i) = 3(4 + 2i)3 + 23(4 + 2i)2 + 52 (4 + 2 i) + 20 = 3(16 + 88i) – 23(12 + 16i) + 208 + 104i + 20 = 48 + 26 4i – 27 2766 – 36 3688i +  208  + 104 i + 20   =0 (4 + 2i)2 = 16 + 16i + 4i2 = 12 + 16i  (4 + 2i)3 = (12 + 16i) (4 + 2i) = 48 + 24i + 64i + 32i2  = 16 + 88i  Since f (4 (4 + 2i) = 0 ⇒ 4 + 2i is a root of the equation. Since all the coefficients are real, complex roots occur in conjugate pairs. Therefore 4 – 2i is a root of f (  zz) = 0 A quadratic factor of f (  zz) is: (  zz – (4 + 2i)) (  zz – (4 – 2i)) = z2 – (4 – 2i) z – (4 + 2i) z  z + (16 + 4)

=

Unit 2 Answers: Chapter 1

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Page 11 of 58 

=  z 2 − 4z + 2iz   − 4 z − 2iz   + 20   = z2 – 8 z + 20  Now: 3 z3 – 23 z2 + 52 z + 20 = ( z2 – 8 z + 20) (3 z + 1) ∴  zz2 – 8 z + 20 = 0, 3 z + 1 = 0 1  z = 4 + 2 i or 4 – 2i or   −   3 3

13 

2

4 z  – 7 z  + 6 z + 2 = 0 Let f (  zz) = 4 z3 – 7 z2 + 6 z + 2  f (1 (1 + i) = 4(1 + i)3 – 7(1 + i)2 + 6(1 + i) + 2 = 4(1 + i)(1 + i)2 – 7(1+ 2i + i2) + 6 + 6i + 2 = 4(1 + i)(2i) – 14i + 6i + 8 = 8i +   8i  2 – 14i +  6i  + 8   =0 Since f  (1  (1 + i) = 0 ⇒ 1 + i is a root of f (  zz) = 0 Since all coefficients are real, complex roots occur in conjugate pairs ∴ 1 – i is also a root A quadratic factor is: (  zz – (1 + i)) (  zz – (1 – i)) = z2 – (1 – i)  zz – (1 + i)  zz + 12 + 12  =  z 2  – z + iz  –  z –  iz +  2   = z2 – 2 z + 2

4 z3 –2 7 z2 + 6 z + 2 = ( z2 – 2 z + 2) (4 z + 1) ∴ (  zz  – 2 z + 2) (4 z + 1) = 0 ⇒  zz2 – 2 z + 2 = 0, 4 z + 1 = 0 1  z = 1 + i or 1 – i or −   4 Since 3 – 2i is a root and all coefficients are real, 3 + 2 i is also a root 14  A quadratic factor is: (  zz – (3 – 2i)) (  zz – (3 + 2i)) = z2 – (3 + 2i) z – (3 – 2i)  zz + (3 – 2i) (3 + 2i) = z 2 − 3z − 2iz   − 3 z + 2iz   + 9 + 4   = z2 – 6 z + 13  f (  zz) = z3 – 8 z2 + 25 z – 26 = (  zz2 – 6 z + 13) (  zz – 2) ∴  zz = 3 – 2i or 3 + 2i or 2  z3 – 5 z2 + 8 z – 6 = 0 15   z = 3, (3)3 – 5(3)2 + 8(3) – 6 = 27 – 45 + 24 – 6 = 51 – 51 = 0 ∴  zz – 3 is a factor  z 2 − 2 z + 2  z − 3 z 3 − 5 z 2 + 8 z − 6  z 3 − 3z 2

− 2 z 2 + 8 z − 6   −2 z 2 + 6 z 2 z − 6 2 z − 6 0

Unit 2 Answers: Chapter 1

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Page 12 of 58 

∴ (  zz – 3) (  zz2 – 2 z + 2) = 0 ⇒  zz = 3, z2 – 2 z + 2 = 0  2 ± 4 − 8   2 2 ± 2i = =1± i   2 ∴  zz = 3 or 1 + i or 1 – i 

 z =

Try these 1.3

(a) (i)

| 5 + i |= 52 + 12 = 26       1  arg(5 + i) =  ttaan −1   = 0.197 rad   5

(ii) | 3 − i |= 2  

arg( 3 − 1) = tan −1 ((0 0) = 0   (iii) | − 3 − i |=

2

( − 3 ) + (−1)2 = 4 = 2  

arg (− 3 − i) = −π  + tan −1     −1  = − 5π   6 − 3 (iv) | − 3 + i |= 3 + 1 = 2  

5π   1   +π=    6 − 3

arg(− 3 + i) =  tan −1  (b) (i)

| 3 + 4i |= 32 + 42 = 5       4  arg(3 + 4i) =  tan −1   = 0.927 rad   3

(ii) | 2 − 4i |= 4 + 16 = 20  

    4  arg(2 − 4i) =  ttaan −1  −  = −1.107 rad    2 (iii) | −2 + 5i |= (−2) 2 + 52 = 29       5  arg(−2 + 5i ) = π  + tan −1   = 1.951 rad    −2  (iv) | −4 − 7i |= 16 + 4 49 9 = 65     −7  arg(−4 − 7i) = −π  + tan −1   = −2.09 rad    −4  Exercise 1C 1 

(a)

2 + 5i  =

22 + 52 = 29  

(b)

3 + 7 i  =

32 + 72 = 58  

Unit 2 Answers: Chapter 1

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Page 13 of 58 

(c)

−1− 4 i  = ( −1)2 + (−4)2 = 17  

(d)

−1+ 2 i  = ( −1)2 + 22 = 5  

(e)

cosθ + i 2 sin θ   =

cos2 θ + (2 ssiin θ )2  

= cos 2 θ + 4 ssiin 2 θ    = cos2 θ + 4 (1 − cos2 θ )   = 4 − 3 co cos 2  θ    2 

    4  (a) arg (2 + 4i) = tan  −1      2 = 1.107 radians

  −1 (b) arg (3 – i) = tan  −1      3 = –0.322 radians

(c) arg (–1 + 2i) = tan – 1 (–2) + π  = 2.034 radians.

Unit 2 Answers: Chapter 1

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Page 14 of 58 

  −2  (d) arg (–4 – 2i) = tan −1   − π    −4  = –2.678 radians

3 

We need the modulus and argument of each number: (a) r = | 2 − 3i | = (2) 2 + (− 3 3))2 = 7      −  3   2  = − 0.714 radians    

θ  = arg (2 −  3i ) = ta tan −1 

Substituting into r  [cos  [cosθ   + + i sinθ ] and reiθ : 2 − 3 i = 7 [cos ( − 0.714) + i sin ( − 0. 0.714)]  

= 7 [cos (0.714) − i sin (0.714)]     i = 7  e − 0.714  

(b)

r

= | − 3 + 2i | =

2

( − 3 ) + 22 = 7      2   +π   3

θ = arg ( − 3  + 2i ) = tan −1  −

= 2.285 radians

Unit 2 Answers: Chapter 1

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∴ − 3 + 2i = 7 [cos (2.285) + i sin (2.285)]     i   = 7  e2.285

2 2 r = |1 − i | = (1) + (− 1) =

(c)

2 

π   1    4  1   π  π  ∴ 1 − i = 2 cos  −  + i sin  −      4  4     θ  = arg (1 − i) = tan −1  −  = −

= 2e  

π − i 4

  9

4

π π  (a)  cos 3 + i sin 3    9π 9π = cos + i sin   3 3 = cos 3π + i sin 3π  = –1 + 0i 

10

  2π 2π    (b)  2  cos + i sin 5 5     20π 20π   = 210  cos + i sin    5 5   = 210 [cos 4π + i sin 4π] = 210 + 0i = 1024 6

π π + (c)  cos 18 i sin 18    

= cos = cos =

6π 18

π 3

+ i sin

6π 18

 

π + i sin   3

1 3 i  + 2 2 8

π π  (d)  cos + i sin    2 2  8π 8π = cos + i sin   2 2 = cos 4 π + i sin 4 π  

5

= 1 + 0i  (a) Let us write 1 + i in the form r  (cos  (cosθ   + + i sinθ ). ).

Unit 2 Answers: Chapter 1

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Then use de Moivre’s theorem. 1 +  i  = 2   arg (1 + i) = tan –1 (1) =

π 4

 

π π  ∴ 1 + i = 2  cos + i sin    4 4  20

  π π   Now (1 + i ) =  2  co cos + i sin     4 4    20

20π 20π   = ( 2 ) 20  co + i sin cos   4 4   = 210 (cos 5 π + i sin 5π)   = 1024 (–1 + 0i) = –1024 + 0i

(using de Moivre’s theorem)

(b) | 3 − 3i | = (3)2 + (− 3 ) 2   = 9 + 3 = 12  

    3  arg (3 − 3i ) =  tan − 1  −    3   π =−   6   −π   −π  ∴ 3 − 3i = 12  cos  + i sin       6   6   12

(   )

 Now 3 − 3i 12

=

( 1  2 )

=

(

12

12

   −π   −π   =  12 cos   + i sin        6     6 

  12π   −12π   − + c o s i s i n        6  6   

12

)   [cos (− 2π) + i sin (− 2π)]  

= 2 985 984 + 0i  (c) | − 3 + i | =

(

− 3

)

2

+1 = 2  

    1  arg − 3 + i =  t an −1  − + π   3 

(   )

5π = 6   Unit 2 Answers: Chapter 1

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5π 5π   ∴ − 3 + i = 2  cos + i sin   6 6   9

  5π 5π   Now ( − 3 + i ) =  2  cos + i sin    6 6     45π  45π  = 29 cos  + i sin      6 6       9

  15π   15π   = = 29  cos   + i sin      2   2   = 29(0 – i) = 0 – 512i  (d) | 1 − i | =   (1)2 + (− 1) 2 = 2   π   1  arg (1 − i ) = tan −1  −  = −    1 4

  − π  π  ∴ (1 − i) = 2 cos   + i sin  −      4    4  5    π  π   5 (1 − i ) =  2 cos  −  + i sin  −       4      4   5π   −5π     = ( 2 )5 cos  −  + i sin   4     4

Unit 2 Answers: Chapter 1

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  2 2  = ( 2 )5  − + i   2 2   = –4 + 4i  −3

6 7 

 cos π + i sin π  = cos  −3π  + i sin  − 3π         6 6    6   6  = – ide   Moivre’s theorem: By cos 4θ   + i sin 4θ  =  = (cos θ   + i sin θ )4  = cos4θ   + 4C1 cos3θ   ((i sinθ ) + 4C2 cos2θ   ((i sinθ )2  + 4C3 cos θ   ((i sin θ )3 + (i sin θ )4  ⇒ cos 4θ   + i sin 4θ   = cos4θ   + i 4 cos3θ  sin  sinθ   – 6 cos2θ  sin  sin2θ   – i 4 cos θ  sin  sin3θ   + sin4θ   Equating real and imaginary parts: cos 4θ   = cos4θ   – 6 cos2θ  sin  sin2θ   + sin4θ   sin 4θ   = 4 cos3θ  sin  sin θ   – 4 cos θ  sin  sin3θ . So cos4θ   = cos4θ   – 6 cos2θ  sin  sin2θ   + sin4θ   = cos4θ   – 6 cos2θ  (1  (1 – cos2θ ) + (1 – cos2θ )2  = cos4θ   – 6 cos2θ   + 6 cos4θ  +  + 1 – 2 cos 2θ   + cos4θ   = 8 cos4θ   – 8 cos2θ   + 1. sin 4θ   = 4 cos3θ  sin  sin θ   – 4 cos θ  sin  sin3θ   3

8 

2

= 4 sin θ  (cos  (cos3θ   – cos θ  sin  sin θ ). ). = 4 sin θ  [cos  [cos θ   – cos θ  (1  (1 – cos2θ )] )] 3 = 4 sin θ  (2  (2 cos θ   – cos θ ) By de Moivre’s theorem cos 7θ   + i sin 7θ   = (cos θ   + i sin θ )7  = cos7θ   + 7C1 cos6θ   ((i sin θ ) + 7C2 cos5θ   ((i sin θ )2  + 7C3 cos4θ   ((i sin θ )3 + 7C4 cos3θ   ((i sin θ )4  + 7C5 cos2θ   ((i sin θ )5 + 7C6 cos θ   ((i sin θ )6 + (i sin θ )7  ⇒ cos 7θ   + i sin 7θ  =  = cos7θ   + i 7 cos6θ  sin  sin θ   – 21 cos5θ  sin  sin2θ   – i 35 cos4θ  sin  sin3θ   + 35 cos3θ  sin  sin4θ   + i 21 cos2θ  sin  sin5θ   – 7 cos θ  sin  sin6θ   – i sin7θ   Equating real parts: cos 7θ   = cos7θ   – 21 cos5θ  sin  sin2θ   + 35 cos3θ  sin  sin4θ   – 7 cos θ  sin  sin6θ   = cos7θ   – 21 cos5θ  (1  (1 – cos2θ ) + 35 cos3θ  (1  (1 – cos2θ )2 – 7 cos θ  (1  (1 – cos2θ )3   (1 – 2 cos2θ   + cos4θ ) –  + 35 cos3θ  (1 = cos7θ   – 21 cos5θ   + 21 cos7θ  + 2

4

6

7 cos θ  (1  (1 – 3 cos θ   + 3cos θ  –  – cos θ ) = 64 cos7θ   – 112 cos5θ   + 56 cos3θ   – 7 cos θ   9  By de Moivre’s theorem: cos 3θ   + i sin 3θ   = (cos θ   + + i sin θ )3  = cos3θ   + 3 cos2θ   ((i sin θ ) + 3 cos θ   ((i sin θ )2 + (i sin θ )3  = cos3θ   + i 3 cos2θ  sin  sin θ   – 3 cos θ  sin  sin2θ   – i sin3θ   = cos3θ   – 3 cos θ  sin  sin2θ   + i (3 cos2θ  sin  sin θ   – sin3θ ) Equating real and imaginary parts: cos 3θ   = cos3θ  –  – 3 cos θ  sin  sin2θ   sin 3θ   = 3 cos2θ  sin  sin θ   – sin3θ    sin3θ   Now tan3θ  =   cos3θ  3 co cos2 θ sin θ − sin 3 θ    = 3 cos θ − 3 ccoosθ sin 2 θ  Dividing numerator and denominator by cos 3θ  

Unit 2 Answers: Chapter 1

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3 cos 2 θ sin θ sin 3 θ  − 3 θ c o s cos3 θ    tan3θ  = cos3 θ 3 cosθ sin 2 θ  − cos3 θ cos3 θ  3 ta t an θ − tan 3 θ    = 1 − 3 t an an 2 θ  10  By de Moivre’s theorem cos 5θ   + i sin 5θ   = (cos θ   + + i sin θ )5  = cos5θ   + 5C1 cos4θ   ((i sin θ ) + 5C2 cos3θ   ((i sin θ )2 + 5C3 cos2θ   ((i sin θ )3 + 5C4 cos θ   ((i sinθ )4 + (i sinθ )5  = cos5θ   + i 5 cos4θ  sin  sin θ   – 10 cos3θ  sin  sin2θ   – i 10 cos2θ  sin  sin3θ   + 5 cos θ  sin  sin4θ   + sin5θ   Equating real imaginary parts: cos 5θ  =  = cos5θ   – 10 cos3θ  sin  sin2θ   + 5 cosθ  sin  sin4θ   sin 5θ  =  = 5 cos4θ  sin  sinθ   – 10 cos2θ  sin  sin3θ   + sin5θ   sin 5θ 5 cco os4 θ sin θ − 10 cos2 θ sin 3 θ + sin 5 θ      Now tan5θ  = = cos5θ   c  os5 θ − 10 cos3 θ sin 2 θ + 5 cco os θ sin 4 θ   Dividing numerator and denominator by cos 5θ   5 tan θ − 10 tan 3 θ + tan 5 θ    tan 5θ   = 1 − 10 tan 2 θ + 5 tan 4 θ  sin 5θ 5 cos4 θ sin θ − 10 cos2 θ sin 3 θ + sin 5 θ     = sin θ sin θ  = 5 cos4θ   – 10 cos2θ  sin  sin2θ   + sin4θ   = 5 cos4θ   – 10 cos2θ  (1  (1 – cos2θ ) + (1 – cos2θ )2  = 5 cos4θ   – 10 cos2θ   + 10 cos4θ   + 1 – 2 cos2θ   + cos4θ   = 16 cos4θ   – 12 cos2θ   + 1 cos 3θ + i sin 3θ  = (cos 3θ + i sin 3θ ) (cos 5θ + i sin 5θ )−1    12  cos 5θ + i sin 5θ  = (cos 3θ   + i sin 3θ ) [cos(–5θ ) + i sin(–5θ )] )] = (cos 3θ   + i sin 3θ ) (cos 5θ   – i sin 5θ ) = cos 3θ  cos  cos 5θ   – i cos 3θ  sin  sin 5θ   + i sin 3θ  cos  cos 5θ   – i2 sin 3θ  sin  sin 5θ   = (cos 5θ  cos  cos 3θ   + sin 5θ  sin  sin 3θ ) + i (cos 5θ  sin  sin 3θ   – sin 5θ  cos  cos 3θ ) = cos(5θ  –  – 3θ ) + i sin(3θ  –  – 5θ ) = cos 2θ   – – i sin 2θ   Alternatively cos 3θ + i sin 3θ (cosθ + i sin θ )3     = cos 5θ + i sin 5θ (cos θ + i sin θ )5   11 

= (cosθ + i sin θ ) 3−5   = (cosθ + i sin θ )  −2   cos 3θ + i sin 3θ  So   = cos(−2θ ) + i sin( −2θ )   cos 5θ + i sin 5θ  = cos 2θ − i sin 2θ    sin4θ    13  tan4θ  = cos4θ  By de Moivre’s theorem: 4

cos 4θ   + i sin 4θ   = (cos θ   + + i sin θ )   = cos4θ   + 4C1 cos3θ   ((i sin θ ) + 4C2 cos2θ   ((i sin θ )2 + 4C3 cos θ   ((i sin θ )3 + (i sin θ )4 

Unit 2 Answers: Chapter 1

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= cos4θ   + i 4 cos3θ  sin  sin θ   – 6 cos2θ  sin  sin2θ   – i 4 cos θ  sin  sin3θ   + sin4θ   Equating real and imaginary parts: cos 4θ   = cos4θ   – 6 cos2θ  sin  sin2θ   + sin4θ   sin 4θ   = 4 cos3θ  sin  sin θ   – 4 cos θ  sin  sin3θ    sin  sin 4θ    tan4θ  = cos4θ  3

3

= 4 c4os θ sin θ2 − 4 co2 s θ sin θ4    cos θ − 6 cos θ sin θ + sin θ   Dividing numerator and denominator by cos 4θ   4 tan θ − 4 tan 3 θ    tan 4θ   = 1 − 6 tan 2 θ + tan 4 θ    π 5π 9π 1133π   , , , 4 4 4 4

Let tan 4θ = 1 ⇒ 4θ  =

  π 5π 9π 13π   , , , 16 1166 16 16 4 tan θ − 4 tan 3 θ  = 1  1 − 6 tan 2 θ + tan 4 θ 

θ  =

4 tanθ   – 4 tan3θ  =  = 1 – 6 tan 2θ   + tan4θ   4

3

2

⇒ t   = tan  = tan θ  θ    + 4 tan  θ   – 6 tan  θ   – 4 tan θ   + 1 = 0 4 3 2 t   + 4t   – 6t   – 4t  +  + 1 = 0  nπ  t = tan  5, 9, 9, 13    , n = 1, 5,

 16  5 14  (a) (cos 3θ   + i sin 3θ ) (co sθ   + i sin θ )   = (cos 3θ   + i sin 3θ ) (cos 5θ   + i sin 5θ ) = cos 3θ  cos  cos 5θ   + i cos 3θ  sin  sin 5θ   + i sin 3θ  cos  cos 5θ   + i2 sin 3θ  sin  sin 5θ   = (cos 3θ  cos  cos 5θ   – sin 3θ  sin  sin 5θ ) + i (cos 3θ  sin  sin 5θ   + sin 3θ  cos  cos 5θ ) = cos(3θ  +  + 5θ ) + i sin(3θ  +  + 5θ ) = cos 8θ   + i sin 8θ   (b) (cos 2θ   + i sin 2θ ) (cos θ   + i sin θ )7  = (cos 2θ   + i sin 2θ ) (cos 7θ   + i sin 7θ ) = cos 2θ  cos  cos 7θ   + i cos 2θ  sin  sin 7θ   + i sin 2θ  cos  cos 7θ   + i2 sin 2θ  sin  sin 7θ   = (cos 2θ  cos  cos 7θ   – sin 2θ  sin  sin 7θ ) + i (cos 2θ  sin  sin 7θ   + sin 2θ  sin  sin 7θ ) = cos(2θ  +  + 7θ ) + i sin(2θ  +  + 7θ ) = cos 9θ   + + i sin 9θ   cosθ − i sin θ  (c) = (cosθ − i sin θ ) (cos (−4θ ) + i sin ( −4θ )) )) −1    cos 4θ − i sin 4θ  = (cos θ   – i sin θ ) (cos 4θ   + i sin 4θ )  sin 4θ   = cos θ  cos  cos 4θ   + i cos θ  sin  sin 4θ   – i sin θ  cos  cos 4θ   – i2 sin θ  sin = cos θ  cos  cos 4θ   + sin θ  sin  sin 4θ   + i (cos θ  sin  sin 4θ   – sin θ  cos  cos 4θ ) = cos(4θ   – – θ ) + i sin (– θ   + 4θ ) θ + = cos 3θ   + i sin 3θ  

Unit 2 Answers: Chapter 1

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Exercise 1D 1 

(a)

(b)  z − i = 4   Circle centre (0, 1) radius 4

(c)  z + 4 = 2 ⇒ z − ( − 4) = 2   Circle centre (– 4, 0) radius 2

Unit 2 Answers: Chapter 1

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(d)  z − 1 + 2i = 5 ⇒ z − (1 − 2 i ) = 5   Circle centre (1, –2) radius 5

(e)  z + 1 + 3i = 6  

⇒  z − ( − 1 − 3i ) = 6   Circle centre (–1, –3), radius 6

Unit 2 Answers: Chapter 1

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(f)

 z

+ 2 − 4i = 7  

⇒   z − (− 2 + 4i) = 7   Circle centre (–2, 4), radius 7

2 

(a)  z − 1 − i = z − 1 + 2 i  

⇒   z − (1 + i ) =

z

− (1 − 2 i )  

Locus of z is the perpendicular bisector of the line joining (1, 1) to (1, –2)

Unit 2 Answers: Chapter 1

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(b)  z − 3 + i = z + 1 + 2 i  

⇒  z − (3 − i ) = z − ( −1 − 2i )   Locus of z is the perpendicular bisector of the line joining (3, –1) to (–1, –2)

(c)  z − 3i = z   Locus of z is the perpendicular bisector of the line joining (0, 3) to (0, 0)

Unit 2 Answers: Chapter 1

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Page 25 of 58 

(d)  z + 2 = z − 2    z − ( − 2) = z − 2  

Locus of z is the perpendicular bisector of the line joining (–2, 0) to (2, 0)

(e)

| z + 1 + 4i | = 1  | z − 1 − 2i | ⇒  z − (− 1 − 4i) = z − (1 + 2i)   Locus of z is the perpendicular bisector of the line joining (–1, –4) to (1, 2)

Unit 2 Answers: Chapter 1

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Page 26 of 58 

(f)  z − 1 − 7i = z + 1 + i  

⇒   z − (1 + 7i) = z − (− 1 − i)   Locus of z is the perpendicular bisector of the line joining (1, 7) to (–1, –1)

3 

(a) arg (  zz) =

π

  2 Locus of z is a half-line starting at (0, 0) excluding (0, 0) and making an angle of

π 2

 radians with the positive real axis

Unit 2 Answers: Chapter 1

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Page 27 of 58 

(b) arg (  zz) =

−π

  6 Locus of z is a half-line starting at (0, 0) excluding (0, 0) and making an angle of

−π 6

 radians with the positive real axis

(c) arg (  zz – 1) =

π

  4 Locus of z is a half-line starting at (1, 0) excluding (1, 0) and making an angle of

π 4

 radians with the positive real axis

(d) arg (  zz – i) =

π

  12 Locus of z is a half-line starting at (0, 1) excluding (0, 1) and making an angle of

π 12  radians with the positive real axis

Unit 2 Answers: Chapter 1

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Page 28 of 58 

3π   4 3π arg (  zz – (3 – 2i)) =   4 Locus of z is a half-line starting at (3, –2) excluding (3, –2) and making an angle 3π  radians with the positive real axis of 4

(e) arg (  zz – 3 + 2 i) =

(f) arg (  zz – 3 – 4i) = −2π   3

  −2 π   ⇒ arg ( z − (3 + 4i)) =

3 Locus of z is a half-line starting at (3, 4) excluding (3, 4) and making an angle of −2 π  radians with the positive real axis 3

Unit 2 Answers: Chapter 1

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Page 29 of 58 

4 

(a)  z = 1 + 2 i + λ  (1  (1 – 3i), λ ∈¡   Locus of z is a line passing through (1, 2) and parallel to 1 – 3 i 

(b)  z = 1 – 2i + λ  (3  (3 + 2i), λ ∈¡   Locus of z is a line passing through (1, –2) and parallel to 3 + 2 i 

Unit 2 Answers: Chapter 1

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(c)  z = i + λ  (4  (4 + i), λ ∈¡   Locus of z is a line passing through (0, 1) and parallel to 4 + i 

(d)  z = 3 – 2i + λ  (5  (5 + 2i), λ ∈¡   Locus of z is a line passing through (3, –2) and parallel to (5 + 2 i)

(e)  z = 1 – 4i + λ  (–1  (–1 – 3i), λ ∈¡   Locus of z is a line passing through (1, –4) and parallel to –1 – 3 i 

Unit 2 Answers: Chapter 1

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(f)  z = 2 + λ  (4  (4 + 2i), λ ∈¡   Locus of z is a line passing through (2, 0) and parallel to 4 + 2 i 

5 

(a)  z − 2 ≤ 3  

Unit 2 Answers: Chapter 1

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Page 32 of 58 

(b)  z − 3 < z − i  

(c)  z − 3 ≤ 2  

(d)  z − 2 i ≤ z + 3 − i  

Unit 2 Answers: Chapter 1

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Page 33 of 58 

(e) arg (  zz – i) ≥ 

π 4

 

(f) arg (  zz – 1 + 3 i) ≤ 

2π   3

Unit 2 Answers: Chapter 1

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Page 34 of 58 

6 

(a) | z − 1 + 2i | = 2 ⇒ | z − (1 − 2i ) | = 2   Locus of z is a circle with centre (1, –2) and radius 2

(b)   z |z + 3 + 2i| =   z |z – 1 – i| ⇒ |  zz – (–3 – 2i)| = 12 – (1 + i)| Locus of z is the perpendicular bisector of the line joining (–3, –2) to (1, 1)

(c) arg (  zz – 1 + i) =

π

 ⇒ arg (  zz – (1 – i)) =

π

  3 3 Locus of z is a half-line starting at (1, –1) excluding (1, –1) and making an

π angle of 3  radians with the positive real axis

Unit 2 Answers: Chapter 1

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Page 35 of 58 

3π 3π    ⇒ arg (  zz – (2 + 3i)) = 4 4 Locus of z is a half-line starting at (2, 3) excluding (2, 3) and making an 3π  radians with the positive real axis angle of 4

(d) arg (  zz – 2 – 3i) =

7 

(a)

 z + 2 + 3i

= 5 ⇒ z − (− 2 − 3i) = 5  

Circle centre (–2, –3) radius 5

Unit 2 Answers: Chapter 1

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Page 36 of 58 

−2 π

−2 π

  3 3 Locus of z is a half-line starting at (2, 2) excluding (2, 2) and making an −2π angle of  radians with the positive real axis 3

(b) arg (  zz – 2 − 2i)) =

 ⟹ arg (  zz – (2 + 2i)) =

(c)  z − 3 − i = z + 4 + 2 i ⇒ z − (3 + i ) = z − ( − 4 − 2i )   Locus of z is the perpendicular bisector of the line joining (3, 1) to (–4, –2)

(d)  z = (1 + i) + λ  (–3  (–3 + 5i), λ ∈¡   Locus of z is a straight line passing through (1, 1) and parallel to –3 + 5 i 

Unit 2 Answers: Chapter 1

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Page 37 of 58 

8 

(a)  z − 2 = 3   Circle centre (2, 0), radius 3  z − 2 − 2i = z ⇒ z − (2 + 2i ) = z   Perpendicular bisector of the line joining (2, 2) to (0, 0)

π b =   4 3 π 3 2 b = 3 sin =   4 2

sin

π

a

cos 4 = 3  

Unit 2 Answers: Chapter 1

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Page 38 of 58 

a = 3 cos

π 3 2   = 4 2

 3 2 3 2 i  ∴ point of intersection =  2 − + 2   2   3 2  = + c 2 2   −3 2 d   = =   2  3 2  2  Next point of intersection is  2 +  − 3 2  i   2     9 

 z

− 3 + i = z + 1 + 2i  

Let z = x + iy ⇒  x + iy − 3 + i = x + iy + 1 + 2 i   ⇒ ( x − 3) + i ( y + 1) = (x + 1) + i ( y + 2)  

⇒ ( x − 3)2 + ( y + 1)2 = ( x + 1)2 + ( y + 2)2   ⇒ ( x − 3)2 + ( y + 1)2 = (x + 1)2 + ( y + 2)2  

⇒  x 2 − 6 x  + 9 +

y2

+ 2 y  + 1 =

x2

+ 2 x  + 1 +

y 2 

+ 4 y + 4  

8 x + 2 y – 5 = 0 The Cartesian equation of the locus is 2 y + 8 x – 5 = 0 10   z − 2 + 3i = 4   Let z = x + iy ⇒  x + iy − 2 + 3i = 4  

⇒ ( x − 2) + i ( y + 3) = 4   ⇒ ( x − 2)2 + ( y + 3)2 = 4   ∴ (  xx – 2)2 + (  yy + 3)2 = 42 Locus of z is a circle with centre (2, –3) and radius 4 The Cartesian equation is x2 – 4 x + y2 + 6 y – 3 = 0 2π 11  arg (  zz – 3 – 4i) = − 3   Let z = x + iy 2π ∴ arg (  xx + iy – 3 – 4i) = −   3   − 2π arg ( x − 3 + i ( y − 4)) =   3     y  − 4  −2π   ⇒ tan −1  =   x − 3  3  y − 4  x − 3

= 3 

 y – 4 =

3 (  xx – 3)

 y = 3x − 3 3 + 4, 4, x > 3  

Locus of z is a line with gradient

Unit 2 Answers: Chapter 1

3  and x > 3

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Page 39 of 58 

12  (a)

 z + 2 = z − 1  

Perpendicular bisector of the line joining j oining (–2, 0) to (1, 0)

(b) arg (  zz – i) =

π 4

 

Half-line starting at (0, 1) excluding (0, 1) and making an angle of

π 4

 radians

with the positive real axis

(c)

 z − 2 + 5i

= 3 ⇒ z − (2 − 5i) = 3  

Circle centre (2, –5) radius 3

Unit 2 Answers: Chapter 1

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Page 40 of 58 

13   z − (3 + 3i ) = z  

Perpendicular bisector of the line joining (3, 3) to (0, 0)  z − 3 = 4   Circle centre (3, 0) radius 4

sin cos

π 4

π

b

= ⇒ b = 4 sin 4

π 4 2 = =2 2   4

2

π = sin = 2 2  

4 4 a = 3 − 2 2  

d  = 2 2   c = 3 + 2 2  

Unit 2 Answers: Chapter 1

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Page 41 of 58 

Points of intersection: (3 − 2 2) 2 ) + (2 2) 2 )i   and (3 + 2 2) 2 ) − (2 2) 2 )i .

− 4 − 2i = 1   ⇒  z − (4 + 2 i ) = 1  

14   z

Circle centre (4, 2) radius 1

Length of OC =   42 + 22 = 20    z1  has the smallest argument and  z2  has the largest argument

sin α =

1 20

⇒ α  = 0.226  

2 ⇒ β = 0.464   4 arg (  zz1) = β   – – α  =  = 0.464 – 0.226 = 0.238 radians tan β =

 z 1 =  ( 20 )2 − 12  

= 19   (0.238) + i sin(0.238)]   ∴ z 1 = 19 [cos (0. = 4.236 + 1.028 i  arg (  zz2) = β   + + α  =  = 0.464 + 0.226 = 0.69  z 2 =  19  

∴ z2 = 19  [cos 0.69 + i sin 0.69] = 3.362 + 2.775i  The complex number with the smallest argument is 4.236 + 1.028 i  The complex number with the largest argument 3.362 + 2.775 i 15   z − 2i = z − 4   Perpendicular bisector of the line joining (0, 2) to (4, 0)

Unit 2 Answers: Chapter 1

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Page 42 of 58 

arg (  zz – 1) =

π 4

 

Half-line starting at (1, 0) excluding (1,0) and making an angle of

π 4

radians with the

 positive real axis

Point of intersection is 2 + i  1 16   z − 1 − i =   2 1  z − (1 + i ) =   2 1 Circle centre (1, 1) radius   2

Unit 2 Answers: Chapter 1

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Page 43 of 58 

1 1 ⇒ θ  = 0.361 radians sin θ = 2 = 2 2 2 π  1 tan β = 1 ⇒ β  = 4   α   = = β   – – θ   = =

π

 – 0.361 = 0.424 radians 4 Smallest argument = 0.424 radians Largest argument =

π 4

+ 0.424 = 1.209 radians

Review exercise 1 1 

(a)

(b)

−1 + 2i   3+i −1 + 2i 3 − i = × 3+i 3−i   −3 + i + 6i − 2i 2   = 9+1 −1 + 7i −1 7 = = + i  10 10 10 10

− 5 + 12i =   (−5)2 + 12 2   = 13     12  arg(−5 + 12i) = tan −1   + π    −5  = 1.966 radians

Unit 2 Answers: Chapter 1

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Page 44 of 58 

2

5 − 12i   3 + 4i 5 − 12i 3 − 4i   = × 3 + 4i 3 − 4i

=

 z

15 − 20i − 36i + 48i 2   32 + 42  −3333 − 56i   = 25 −33 56

=

=

−

i 

25

3 

25 2 2  −33   −56  13   z |z| =   +  =   5  25   25   −56   −33 56      −1  25  − i  = tan  arg  − π   25 25     −33   25  = –2.103 radians Let ( x + iy) = 16 − 30i  

⇒  ( x + iy)2 = 16 − 30i   ⇒  xx2 – y2 + i (2 xy) = 16 – 30i ⇒  xx2 – y2 = 16 2 xy = –30 From (2) ⇒  y =

 −15  x

[1] [2]

 

Substitute into [1] 2

  −15  = 16    x −    x  2

 x 2 4

−

225  x 2

= 16   2

 x  – 225 = 16 x  x4 – 16 x2 – 225 = 0 (  xx2 – 25) (  xx2 + 9) = 0 2 2  x  = 25, x  = –9 Since x is real x2 = 25, x = ±5

−15 When x = 5,  y = 5 = − 3  

Unit 2 Answers: Chapter 1

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Page 45 of 58 

  −15 = 3  −5

When x = –5,  y =

∴  16 − 30 3 0i = 5 − 3i , –5 + 3i 4 

5

Let  f (  zz) = 2 z3 – 3 z3 + 32 z + 17  f (1 (1 + 4i) = 2(1 + 4i)3 – 3(1 + 4i)2 + 32(1 + 4i) + 17 (1 + 4i)2 = 1 + 8i + 16i2 = –15 + 8i  (1 + 4i)3 = (1 + 4 i) (–15 + 8i) = –15 + 8i – 60i + 32i2 = –47 – 52i ∴  f (1 (1 + 4i) = 2(–47 – 52i) – 3(–15 + 8i) + 32(1 + 4i) + 17 = –94 – 104i + 45 – 24i + 32 + 128i + 17 = –94 + 94 –104i + 104i =0 ⇒ 1 + 4i is a root of the equation. Since roots occur in conjugate pairs for real coefficients 1 – 4i is also a root.  A quadratic factor is (  zz – (1 + 4i)) (  zz – (1 – 4i)) = z2 –  – (1 – 4i) z – (1 + 4i) z + (1 + 4i) (1 – 4i) 2 = z  –  – 2 z + 17.  Now 2 z3 –  – 3 z2 + 32 z + 17 = ( z2  – – 2 z + 17) (az + b) ⇒ a = 2 b = 1 17  + 17) (2 z + 1) = 0 ∴ (b  zz =2 –172 z⇒  1 ⇒  zz = 1 + 4i, 1 – 4i, −   2 2 (a)  z   – – 2 z + 6 = 0

2 ± ( −2)2 − 4(1)(6)    z = 2  2 ± −20   = 2  2 ± 2i 5   = 2 =1± i 5   (b)

(c) (i) (ii)

1 + i 5 = 12 +

( ) 5

2

= 6 

arg (1 + i 5 ) =  tan −1     5  = 1.150 radians  1 

Unit 2 Answers: Chapter 1

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Page 46 of 58 

6 

(

)

 z − −2 + 2 3i ≤ 2  

The locus ooff  z for z − ( −2 + 2 3i ) = 2  is a circle centre (–2, 2 3 ) and radius 2.

(a)

OA = least value of   zz  

(

OC = (−2)2 + 2 3

)

2

= 4 + 12 = 4  

AC = 2   ∴ OA = 4 − 2 = 2   (b) Greatest possible value of arg (  zz) is α    = β  +  + θ . α  =

θ

  2 3    π 2π = arg ( −2 + 2 3i) = tan −1   + π = π −  =   3 3  −2 

tan β =

2 1   1 = ⇒ β =  ta  n −1   = 0.464 radian    2 4 2

  2π 0.4644 = 2.55 2.5588 radi radian anss   ∴α  = + 0.46 3

7

(a)

(

1 − 3i = 12 + − 3

)

2

= 4 =2 

   −  3  π arg 1 − 3i =  t an −1   = −   3  1 

(

)

−

(b)

∴ 1 − 3i = 2e

π 3

i

 

sin α − i cos α = (sin α )2 + ( − cos α )2    

Unit 2 Answers: Chapter 1

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Page 47 of 58 

= sin 2 α + cos2 α    = 1  =1  

   − cos α     α  sin  

arg (sin α − i cos α ) = tan  −1  = tan −1 ( − co  t α )  

  π = tan −1 (− tan( − α ))   2

   π  = tan −1   tan  α −      2   

 π = α −    2

∴ sin α − cos α  = e    ( α − π/ 2) i   (c) 1 + cos 2θ  +  + i sin 2θ     1 + cos 2θ + i sin  2θ = (1 + cos 2θ)2 + sin 2 2θ  

 

= 1 + 2 ccoos 2θ + cos2 2θ + sin 2 2θ    cos 2θ    = 2 + 2 co = 2(1 + cos2θ )     2co os2 θ )   = 2(2c

= 2 cos θ  

     sin2θ      1 + cos 2θ 

arg (1 (1 + cos 2θ + i sin2θ )  = tan −1 

8

2sin n θ cosθ      2si = tan  −1    2  2cos θ   = tan-1(tanθ ) = θ   iθ   ∴1 + cos 2θ  +  + i sin 2θ  = 2 cosθ   ee  f (  zz) = 3 z3 – 16 z2 + 27 z + 26 = 0  f (3 (3 + 2i) = 3(3 + 2i)3 – 16(3 + 2i)2 + 27(3 + 2i) + 26 (3 + 2i)2 = 9 + 12i + 4i2 = 5 + 12i  (3 + 2i)3 = (3 + 2 i) (5 + 12i) = 15 + 36i + 10i + 24i2 = –9 + 46i ∴  f f (  zz) = 3(–9 + 46i) –16(5 + 12i) + 27(3 + 2i) + 26 = –27 + 138i – 80 – 192i + 81 + 54i + 26 = –107 + 107 + 138i – 138i =0 ∴ 3 + 2i is a root of f (  zz) = 0 Since all coefficients are real, 3 – 2 i is also a root A quadratic factor is: ( z  z – (3 + 2i)) (  zz –  (3 – 2i)) = z2  –  –  (3 – 2i) z –  (3 + 2i) z + (3 + 2i) (3 – 2i) 2  z   =  –– 6 z + 13 3 z3 – 16 z2 + 27 z + 26 = ( z2 –  – 6 z + 13)(az + b) Coefficient of z3 ⇒ a = 3 26 = 13b ⇒ b = 2

Unit 2 Answers: Chapter 1

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Page 48 of 58 

∴ (  zz2 –  – 6 z + 13) (3 z + 2) = 0  2 3

−  

 z = 3 + 2 i, 3 – 2i,

9

(a)

 z

= z−4  

Locus is the perpendicular bisector of the line joining (0,0) and (4,0) π

(b) arg(  zz – i) =

4

 

Locus is a half-line starting at (0,1) excluding (0,1) and making an angle of

π

4

 

radians with the positive real axis

Point of intersection is (a ,b) a = 2

tan 

π

4

 = 

c = 2 tan 

c

 

2 π

4

  = 2 

∴ b = 2 + 1 = 3 Point of intersection is 2 + 3i  10 

3 −i = arg

(

( 3)

2

+ (−1)2 = 4 = 2        −1  −π =    3 6

)

3 − i =  t an −1 

 π π  ∴  3 − i = 2 cos  −  + i sin  −     

 6

Unit 2 Answers: Chapter 1

 6 

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Page 49 of 58 

6

⇒( 3 − i)

6

 π π  = 2 cos  −  + i sin  −      6    6

    6π   6π   = 26 cos  −  + i sin  −   de Moivre’s theorem    6    6  = 64 [cos(– π) + i sin(– π)] = –64 + 0i 3 + 4i 3 + 4i 1 + 2i = ×   11  (a) w = 1 − 2i 1 − 2i 1 + 2i 3 + 6i + 4i + 8i 2   = 1+ 4 −5 + 10i = = −1 + 2i   5 (b)

(c) Let the greatest value of arg z = θ    = α  + β   θ  = OC = (−1)2 + 22 = 5   sin β =

  1  1 ⇒ β = s  in −1  = 0.464 radians  5  5     2     −1 

α  = arg(−1 + 2  i) = π + tan  −1 

= = 2.034 radians  = 2.034 + 0.464 = 2.498 radians θ  2 3 12   f (  zz) = 2 z  + z  – 4 z + 15

(

 f 1 +

) (

3

) (

2 i = 2 1 + 2i + 1 + 2 i 2

)

2

− 4 (1 + 2i ) + 15  

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

3

2i − 2i + 4i 2  

= −5 + 2i   ∴  f  (1 + 2i ) = 2 ( −5 + 2i ) + (−1 + 2 2i ) − 4 (1 + 2i ) + 15   = − 10 + 2 2i − 1 + 2 2i − 4 − 4 2i + 15  

=  −15 + 15 15 + 4 2i − 4 2i = 0   ∴  1 + 2i is a root of f ( z ) = 0   Unit 2 Answers: Chapter 1

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Page 50 of 58 

Since all the coefficients are real, 1 − 2i  is also a root. A quadratic factor is

( z − (1 + 2i) ) ( z − (1 − 2i) )   =  z 2 − (1 − 2i ) z − (1 + 2i ) z + (1 + 2i ) (1 − 2i )   =  z 2 − 2 z + 3.   2 z3 + z 2 − 4 z + 15 = (z 2 − 2 z + 3) (az + b)   Coefficient of  z 3 ⇒ 2 = a   15 = 3b ⇒ b = 5 ∴  2 z 3 + z 2 − 4 z + 15 15 = ( z 2 − 2 z + 3) 3 ) (2 z + 5) = 0     −5   ⇒ z = 1 + 2i, 1 − 2i, 2 13 (a)  z 1 =   1−i   z 1  =  (1)2

+ (−1)2 = 2     −π   arg ( z 1 ) =  tan −1 (− 1) 1) =

4  −π −π  ∴   z 1 =  1 − i = 2  cos   + i sin        4   4  8  z 81 =  (1 − i)8

  −π −π   =  2 cos   + i sin        4      4 

  −8π   −8π    cos  4  + i sin  4          4 = 2  [cos (–2π) + i sin (–2π)] = 16 + 0i (b)  z1 z2 =  5 + 12i   =

8

( ) 2

5 + 12i 5 + 12i 1 + i 5 + 5i + 12 1 2i + 12i 2  z2 = = × =   1− i 1− i 1+ i 2 −7 + 17i −7 17 17 = = + i  2 2 2 iθ 

(c)

 + i sinθ   e  = cosθ  + θ θ θ    = 2 cos2 + i   2 si sin co cos   

2

θ = 2 cos    2



2

2

 θ  θ     + i c o s s i n      2  2   

iθ  θ   2  = 2cos   2 e     iθ  θ    = 2e 2  cos   2

14  (a)

Unit 2 Answers: Chapter 1

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Page 51 of 58 

−5 + 12i = a + bi  

(b) Let

⇒ –5 + 12i = (a + bi)2 2

2

⇒ –5 + 12i = a  – b  + i (2ab) ⇒ a2 – b2 = –5 2ab = 12 12 6 From [2] a = =   2b b

[1] [2]

2

 6 Substituting in [1]   − b2 = − 5    b 36 b

2

− b2 + 5 = 0  

36 – b4 + 5b2 = 0 4 2 b  – 5b  – 36 = 0 (b2 – 9) (b2 + 4) = 0 2 2 b  = 9, b  = –4 b = ±3

b ∈  ¡   6 When b = 3, a =

3

When b = −3, a =

= 2  6 = −2   −3

∴  −5 + 12i = − 2 − 3i , 2 + 3i  z 2 + 4 z + 9 − 12i = 0   a = 1, b = 4, c = 9 – 12i

−4 ± 16 − 4(9 − 12i )   2 −4 ± −20 + 48i   = 2

 z =

= −4 ± 4 −5 + 12i   2

Unit 2 Answers: Chapter 1

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Page 52 of 58 

−4 ± 2 −5 + 12i   2  z = − 2 ± −5 + 12i   =

Substitute −5 + 12 12i = 2 + 3i    z = − 2 + (2 + 3i), − 2 −(2 + 3i)    z = 3i, − 4 − 3i   15   z =

eiα 

  1 − eiα  cos α + i sin α  = 1 − cos α − i sin α      cos α + i sin α 1 − cos α + i sin α = × 1 − cos α − i sin α 1 − cos α + i sin α   (cos α + i sin α) (1 − cos α + i sin α)     = (1 − cos α)2 + sin 2 α 

cos α − cos2 α + i sin α cos α + i sin α − i sin α cos α + i 2 sin 2 α   =   1 − 2 cos α + cos2 α + sin 2 α  cos α − (cos2α + sin 2 α ) + i sin α     = 2 − 2 co cos α  cos α − 1 + i sin α  =   2(1 − cos α )   α  α  i 2sin cos

=

− (1 − cos α ) 2 +   2(1 − cos α )   4sin 2

=− 16 

2

α 

 

2

1 1    α  + i  cot      2 2 2

−7 + 8i = (−7)2 + 82 = 113       8  arg (−7 + 8i ) =  tan −1   π = 2.290 radians −  7 ∴ − 7 + 8i = 113 (cos2.2 s2.290 + i sin 2.290)   8

( −7 + 8i )

(

 8

=  113 (cos 2 2..290 + i sin 2. 2.290 )  

8

)

c  os ( 8 × 2.290 ) + i sin ( 8 × 2.290 )    = 163 047 361(cos 18.32 + i sin 18.32) = 140 715 005.6 – 82 363 396.8i  17  (a) w = 4 – 3 i 1 1   w + = 4 − 3i + 4 − 3i w 1 4 + 3i = 4 − 3i + ×   4 − 3i 4 + 3i   4 + 3i i 4 3 = − + 25   =

113

Unit 2 Answers: Chapter 1

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4 3 + i  25 25 104 72 = − i  25 25 (b) 4i = a + bi  

= 4 − 3i +

2

bii )   4i = (a + b 2 4i = a + (2 ab)i + b2 i 2  

4i = ( a 2 − b2 ) + (2ab) i   ⇒ a2 – b2 = 0 2ab = 4   From [2] a =

2 b

[1] [2]

  2

2 Substituting in [1]   − b 2 = 0   b 4 4 – b  = 0 (2 – b2) (2 + b2) = 0 2 2 b  = 2, b  = −2 Since b is real b = ± 2   2 When b = 2, a = = 2  2   −2 =− 2  When b = − 2, a = 2

∴ 4i = 2 + 2i , − 2 − 2i   (c)

Unit 2 Answers: Chapter 1

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18 

Locus of  z − 1

= z + i  is the perpendicular bisector of the line joining (0,1) and (1,0) Locus of  z − (3 − 3i) = 2  is a circle centre (3,-3) radius 2 Let  z = x + iy   Then in Cartesian form the locus of the line is  x + iy − 1 = x + iy + i   2 (  xx – 1)2 + y2 = x2 + ( y  y + 1)

 x 2  – 2 x  + 1 + y 2   =  x 2  +  y 2  + 2 y + 1    yIn=Cartesian –x  form the locus of the circle is  x – 3)2 + (  yy + 3)2 = 22 ( x

Unit 2 Answers: Chapter 1

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Substituting for y 2 (  xx – 3)2 + (–   x + 3)  = 4 2 2  x  – 6 x + 9 + x  – 6 x + 9 = 4 2 x2 – 12 x + 14 = 0 2  x  – 6 x + 7 = 0 6 ± 36 − 28 28    6 ± 8    x = = 2 2 6±2 2 = =3± 2   2 When  x = 3 + 2 , y = − 3 − 2   When  x = 3 − 2 , y = − 3 + 2   The points of intersection are 3 + 2 − 3 + 2   i  and 3 − 2 − 3 − 2 i  

(

19  (a)

) (

)

(

) (

)

(1 – i)15 1 − i = (1)2 + (−1)2 = 2  

 

arg (1 − i) =  tan −1 (−1) 1) =

−π 4

 

  −π   −π   1 − i = 2  cos  4  + i sin  4     15

   −π  −π   (1 − i ) =  2 cos  + i sin       4      4  15  −15π  −15π   = ( 2 ) cos  + i sin      4 4      15

=

de Moivre’s theorem 

 1 1  + i    2   2

15

( ) 2

 

14

14

= ( 2 ) + ( 2 ) 

i 

= 128 + 128i e

(b)

e

2π i 5 3π i 4

2π

=e

5

i−

3π 4

−7 π

i

=e

20

i

 −7π   −7π  = cos  20  + i sin  20   

= 0.454 – 0.891i  cot θ − i (cot θ − i ) (cot θ  − i )   = 20  cot θ + i (cot θ + i ) (cot θ  − i ) cot 2 θ  − 2 cco ot i + i 2 =   cot 2 θ  − i 2 −1 + cot 2 θ − 2 cot θ i   = 1 + cot 2 θ  −1 +  co cot 2 θ − (2 cot θ ) i  = cosec2θ   )) (1 + cot2θ  = = 2 cosec θ 

Unit 2 Answers: Chapter 1

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cos 2θ cosθ  2 −1 + 2 sin θ sin θ  i  = − 1 1 sin 2 θ sin 2 θ  = (–sin2θ  +  + cos2θ  ) – 2 sinθ  cos  cosθ   ii 2 2 = cos θ  –  – sin θ   – – i (sin 2θ ) = cos 2θ   – – i sin 2θ   = cos (–2θ ) + i sin (–2θ ) cot θ  − i = 1  cot θ  + i

 cot θ  − i  = − 2θ    arg   cot θ  + i  cot θ  − i ∴ = e −2θ i   cot θ  + i 21   z − 1 − i = 2 z − 2 + 3i   Let  z = x + iy  

( x + iy − 1 − i ) =  2 ( x +  iy ) − 2 + 3i   ⇒  ( x −  1) +  i ( y −  1) = 2 x  − 2 + i ( y +  3)   2 2 2 2 ⇒  ( x −  1)  + ( y −  1) = 2 ( x −  2)  + ( y +  3)   2 2 2 2 ⇒  ( x −  1)  + ( y −  1) =  2 2 ( x −  2)  + ( y +  3)    ⇒  xx2 – 2 x + 1 + y2 – 2 y + 1 = 4 [ x2 – 4 x + 4 + y2 + 6 y + 9] ⇒ 3 x2 – 14 x + 3 y2 + 26 y + 50 = 0

  14 26 50 x+ + = 0  3 3 y 3 14 26 50  x 2 − x + y2 + y+ =0   3 3 3 2 2 7 13  49 169 50   − + =0   x −  +  y +  − 3 3 9 9 3  x 2 + y 2 −

 x − 7  2 +  y + 13 2 − 68 = 0       3  3 9  2

2 2  x 7   y 13  68  68     − 3  +  + 3  =  9 =  3     

 7 13  The locus of  z  is a circle with centre  , −  and radius 3 3 n 22  (1 + cosα  + i sinα )

68   3

n

 α α α    = 2 cos2 + i   2 sin cos      2 2 2  n n  2 cos α  cos α i sin α    = +  2      2 2 

Unit 2 Answers: Chapter 1

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n

α  α     i 2   = 2 cos      e    2   n

n

nα 

 α   i 2 = 2 cos     e   2 + i sin 5θ   23  cos 5θ   + = (cosθ   + + i sinθ )5 5 = cos θ   + + 5C 1 cos4θ  (i sinθ ) + 5C 2 cos3θ  (i sinθ )2 + 5C 3 cos2θ  (i sinθ )3 + 5C 4 cosθ  (i sinθ )4 + (i sinθ )5  = cos5θ  + i (5 cos4θ  sinθ  ) – 10 cos3θ  sin2θ   – i(10 cos2θ  sin3θ ) + 5 cosθ  sin4θ   + + i sin5θ   Equating real parts: 4 cos 5θ  = cos5θ  –  – 10 cos3θ  sin2θ  +  + 5 cosθ  sin θ   5 = cos θ  –  – 10 cos3θ  (1 – cos2θ  ) + 5 cosθ  (1 – cos2θ )2 = cos5θ  –  + 10 cos5θ  +  + cos4θ )  – 10 cos3θ  +  + 5 cosθ  (1 – 2 cos2θ  + = 16 cos5θ  –  – 20 cos3θ  + 5 cosθ   cos 5θ  =  = 0 5 ∴ 16 cos θ   –  –  20 cos3θ  +  + 5 cosθ  =  = 0 4 2  +5) = 0 ⇒ cosθ  (16 cos θ   –  –  20 cos θ  + 4  = 0 or 16 cos θ   –   + 5 = 0 ⇒ cosθ  = –  20 cos2θ  +   20 ± 400 − 4(1 4(16)(5)   cos2 θ  = 32   20 ± 80   = 32   20 ± 4 5   = 32 5± 5 =   8     1 + co c os 2θ  cos2 θ =   2 1 + cos 2θ  5 ± 5   = 2 8   5 ± 5   1 + cos 2θ  = 4 5 5 1 5 −1= ±   cos 2θ   = ± 4 4 4 4 Since cos 5θ  =  = 0 -1 5θ  = cos (0) n

5θ   =

n

π

2

  π

θ  =

10

 

  5 π 1   = +  10  4 4

⇒  cos2 

(take the + sign because cos is positive for angles in the first quadrant)

Unit 2 Answers: Chapter 1

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π 1+ 5 ⇒  cos   =   4 5 Using the double angle formula     1 + co cos 2θ    cos2 θ  =

2

cos2  π   = 1 + cos ( π/5 )   2  10  1+ 5 +1 4   = 2 1+ 5 + 4 = 8   5+ 5 = 8

Unit 2 Answers: Chapter 1

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