2022 H2 Maths Paper 1 Q1 — Worked Solution

Solution

4 marks

Approach

With two linear equations in $z$ and $w$ , we eliminate one variable by substitution, then rationalise the resulting complex fraction to reach the $a + ib$ form.

iz + 2w = - 1\quad\text{...(1)}

(2 - i)z + iw = 6\quad\text{...(2)}

From (1),

w = \frac{- 1 - iz}{2}

Substituting into (2),

(2 - i)z + i\left( \frac{- 1 - iz}{2} \right) = 6

Multiplying through by $2$ ,

2(2 - i)z + i( - 1 - iz) = 12

\Rightarrow (4 - 2i)z - i - i^{2}z = 12

Since $- i^{2}z = z$ ,

(4 - 2i)z + z - i = 12

\Rightarrow (5 - 2i)z = 12 + i

\Rightarrow z = \frac{12 + i}{5 - 2i}

Multiplying numerator and denominator by the conjugate $5 + 2i$ ,

z = \frac{(12 + i)(5 + 2i)}{(5 - 2i)(5 + 2i)}

= \frac{60 + 24i + 5i + 2i^{2}}{5^{2} + 2^{2}}

= \frac{58 + 29i}{29}

= 2 + i

Substituting back into $w = \frac{- 1 - iz}{2}$ ,

w = \frac{- 1 - i(2 + i)}{2}

= \frac{- 1 - 2i - i^{2}}{2}

= \frac{- 2i}{2}

= - i

To check your answer: in (1),

i(2 + i) + 2( - i) = 2i + i^{2} - 2i = - 1

✓; in (2),

(2 - i)(2 + i) + i( - i) = \left( 4 - i^{2} \right) + \left( - i^{2} \right) = 5 + 1 = 6

✓

$\therefore$ $z = 2 + i$ and $w = - i$ .

2022 H2 Maths Paper 1, Question 1

Solution

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