2024 H2 Maths Paper 2 Q1 — Worked Solution

Part (a)

2 marks

Since $z = - 2 + i$ , we have $z^{\text{*}} = - 2 - i$ . Substituting into $az^{\text{*}} + b$ ,

az^{\text{*}} + b = a( - 2 - i) + b

= ( - 2a + b) - ai

Equating this to $z = - 2 + i$ ,

Real: $- 2a + b = - 2$

Imaginary: $- a = 1 \Rightarrow a = - 1$

Substituting into the real equation: $- 2( - 1) + b = - 2 \Rightarrow b = - 4$ .

To check your answer:

( - 1)( - 2 - i) + ( - 4) = 2 + i - 4 = - 2 + i = z

✓

$\therefore$ $a = - 1$ , $b = - 4$ .

4 marks

Approach

We compute $wz$ and $\frac{w}{z}$ separately, each by hand without a calculator, then subtract.

wz = (1 - 3i)( - 2 + i)

= - 2 + i + 6i - 3i^{2}

= - 2 + 7i + 3

= 1 + 7i

To divide, we multiply numerator and denominator by the conjugate of $z$ :

\frac{w}{z} = \frac{1 - 3i}{- 2 + i} \times \frac{- 2 - i}{- 2 - i}

= \frac{(1 - 3i)( - 2 - i)}{( - 2)^{2} - i^{2}}

= \frac{- 2 - i + 6i + 3i^{2}}{4 + 1}

= \frac{- 5 + 5i}{5}

= - 1 + i

Subtracting,

wz - \frac{w}{z} = (1 + 7i) - ( - 1 + i)

= 2 + 6i

$\therefore$ $wz - \frac{w}{z} = 2 + 6i$ , so $c = 2$ , $d = 6$ .