561 Answers Chapter 3

Selected answers are given here. For a full set of worked solutions, see Specialist Maths Dimensions Units 3 & 4 Solutions and Study Guide.
13
14 a 5 – 3i, 1 + 3i b 6 – 3i, 2 + 3i
Application Tasks
1 a
Im(z)
(1 – π
2
, 0)
Im(z)
(1, π
2
)
(–1, 0)
(1, – π
2
)
π
3
(1 + π
2
, 0)
Re(z)
π
3
11 a
7
213 + 144i
c i -- + 4i ii -------------------------
2
113
–24 + 12i
d i −3 + 1 ii -----------------------
5
–3
27
y
y = x 3 + 9x 2 + 27x + 27
x
4
2
z 2
2 2
13
z 3
z 1 17
y
y = x 3 + 4x 2 – 3x + 8
2 4 6
Re(z)
b
z' Im(z) 3
6
z' 2 4
z 2
–3 2
–18
x
z' 2 1
z 1
z 3
y
–4 –2 2 4 6
Re(z)
y = x 3 – 2x 2 – 5x + 6
2 ( 2 + 2 3i) 3 = −64
1 3
3 2, -- + ------i, – 1 −2,
2 2 2 -- 3
+ ------i, – 1 2 2 -- 3
– ------i,
2
4
5
6–
2
-------------------
4
Im(z)
(3, 3)
1
-- –
2
3
------i
2
–2 1 3
x
y
–1
y = x 3 + 3x 2 + 6x + 4
x
6
(2, –2) (1, –2) (5, –2)
(3, –2)
Im(z)
5
3
(3, –4)
(3, –7)
π
6
π
6
Re(z)
(8, –2)
π
6
b When there is only one x-intercept and it is not a
point of inflection.
Chapter 4
Exercise 4A
1 a 2sec 2 2x
b
7sec 2 7x
7 a
–1 3 5
–1
–2
Im(z)
Re(z)
(2 + i) 3 = 2 + 11i
c
e
g
6sec 2 3x d – 27sec 2 (– 9x)
– 4sec 2 (–
x) f – 36sec 2 4x
1
--sec 2 x
--
5 5
h
3
--sec 2 x
--
4 4
(2 + i) 2 = 3 + 4i
2 + i
Re(z)
d De Moivre’s theorem holds for fractional powers.
8 a cos 3 θ + 3i sinθcos 2 θ – 3 sin 2 θcosθ
– i sin 3 θ
b cos3θ
+ i sin3θ
9 a i 5 + 6i ii
b i 6 + 5i ii
96 + 80i
-------------------
61
–79 + 119i
--------------------------
61
i
k
5
--sec 2 x
– --
7 7
2
-----sec 2 ⎛ x
–-- ⎞ 6
l -----sec
15 ⎝ 3⎠
2 ⎛ 2x
–-----⎞
25 ⎝ 5 ⎠
2 a 3sec3x tan3x
b – 2cosec2x cot2x
c 18 sec2x tan2x
d 10cosec5x cot5x
e – 21 sec7x tan7x
f 4cosec( – x) cot( – x)
g – 35 sec( – 5x) tan( – 5x)
j
12
-----sec 2 2x
– -----
5 5
Answers
575