561 Answers Chapter 3

Answers
Selected answers are given here. For a full set of worked solutions, see Specialist Maths Dimensions Units 3 & 4 Solutions and Study Guide.
f
Im(z)
Re(z + 1 – i) = –4
h
Im(z)
3
Re(z)
–5
Re(z)
–4
2 a
Im(z)
i
Im(z)
Re(z)
5
(–2, 2 )
–4
Im(z) = –4
Re(z)
b
Im(z)
j
Im(z)
5
2
Im(z)
5
2
5
2
Re(z)
–2
Re(z)
c
Im(z)
3 a
Im(z)
4 Im(z – 3i) = 1
Re(z)
π
4
Re(z)
d
Im(z)
b
Im(z)
Re(z)
π
4
Re(z)
–5
Im(z + 2i) = –3
e
Im(z)
c
Im(z)
5 Im(z + 1) = 6
Re(z)
π
6
Re(z)
f
Im(z)
d
Im(z)
Re(z)
Im(z – 2 + 3i) = 0
–3
π
6
Re(z)
g
Im(z)
e
Im(z)
(3, –4)
Re(z)
π
3
Re(z)
568
Specialist Maths Dimensions Units 3 & 4