Power of Maths: Paper 1 – Section 6 Conclusion
Multiplying a complex number z by a scalar k gives a new complex number w = k z, such that 0 = 0 + 0i, w and z are on the same straight line and the distance from 0 to w is k times the distance from 0 to z.
If k > 0, the line joining 0 to w is in the same direction as the line joining 0 to z.
0
If k < 0, the line joining 0 to w is in the opposite direction to the line joining 0 to z. z
z
w
0 w
w = − 1 _
is 1 _
3 z means 0, w and z are collinear. However, the line joining 0 to w is in the opposite direction to the line joining 0 to z and the distance from 0 to w
3 of the distance from 0 to z. EXERCISE 5
1. Simplify the following giving your answer in the form a + bi, a, b ∈R:
(a) 4(3 + 2i) (b) 5(11 + 10i) (c) 7(6 + 2i) (d) −11(2 + 3i) (e) 3(8 − 2i) (f) −5(−11 + i) (g) −5(−4 − 2i) (h) −6(−2i − 4) (i) 1
_ 2 (−2 − 8i)
(j) 1 _
3 (−6 + 9i)
(k) 5(a + bi ), a, b ∈R (l) k (1 + 2i ), k ∈R
(m) k (a + bi ), k, a, b ∈R (n) 3(1 + i) + 7(1 + i) (o) 3(5 − i) − 7(2 + i) (p) − 1
_ 2 (4 + 8i) − 1
_ 4 (4i + 8)
(q) 2i + 3(1 − i) − 6i (r) −4i − 2(3 + 2i) − 5 (s) 5(2 − i) − 6(5 + 7i) − 3(4 + i)
2. If z = 4 + 6i and w = −5 − 3i, evaluate the following in the form a + bi, a, b ∈R:
(a) z + 2w (b) 2z − w (c) 3z + 2w (d) − z − w (e) −2z + 4w
(f) 1 _
2 z − 3w
(g) − 1 _
2 z + 1 _
(h) 3z − w (i) 3
(j) z − w _____
3
_ 2 z + w
2 w
210
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