2021 H2 Maths Paper 1 Q4 — Worked Solution

Part (a)

3 marks

Approach

No calculator is allowed, so we write each trigonometric factor as $e^{i\theta}$ using $\cos\theta + i\sin\theta = e^{i\theta}$ and combine with index laws, rather than expanding the products directly.

z = \frac{\left( \cos\frac{\pi}{16} + i\sin\frac{\pi}{16} \right)^{2}}{\cos\frac{\pi}{8} - i\sin\frac{\pi}{8}}

The numerator is $\left( e^{i\pi/16} \right)^{2} = e^{i\pi/8}$ , and the denominator is $\cos\frac{\pi}{8} - i\sin\frac{\pi}{8} = e^{- i\pi/8}$ , so

z = \frac{e^{i\pi/8}}{e^{- i\pi/8}}

= e^{i\pi/8 - ( - i\pi/8)}

= e^{i\pi/4}

$\therefore$ $|z| = 1$ and $\arg z = \frac{\pi}{4}$ .

Since $z = e^{i\pi/4}$ ,

z^{2} = e^{i\pi/2}

= \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}

= i

$\therefore$ $z^{2} = i$ .

Part (b)(i)

2 marks

\left( \cos\theta + i\sin\theta \right)\left( 1 + \cos\theta - i\sin\theta \right)

= \cos\theta + \cos^{2}\theta - i\cos\theta\sin\theta + i\sin\theta + i\sin\theta\cos\theta - i^{2}\sin^{2}\theta

The terms $- i\cos\theta\sin\theta$ and $\text{+}i\sin\theta\cos\theta$ cancel, and $\cos^{2}\theta + \sin^{2}\theta = 1$ with $i^{2} = - 1$ , leaving

\left( \cos\theta + i\sin\theta \right)\left( 1 + \cos\theta - i\sin\theta \right) = 1 + \cos\theta + i\sin\theta\ \ \left( \text{shown} \right)

Part (b)(ii)

2 marks

Approach

The identity in (b)(i), read with $\theta = \arg z$ , links $z\left( 1 + z^{\text{*}} \right)$ to $1 + z$ . Raising this to the power $4$ turns $(1 + z)^{4}$ into a multiple of $\left( 1 + z^{\text{*}} \right)^{4}$ , so the required sum factors into $\left( 1 + z^{\text{*}} \right)^{4}\left( z^{4} + 1 \right)$ — which is zero, since $z^{4} = - 1$ .

Let $\theta = \arg z = \frac{\pi}{4}$ , so $z = \cos\theta + i\sin\theta$ and, since $|z| = 1$ , $z^{\text{*}} = \cos\theta - i\sin\theta$ . The identity of (b)(i) then reads

z\left( 1 + z^{\text{*}} \right) = 1 + z

Raising both sides to the power $4$ ,

(1 + z)^{4} = z^{4}\left( 1 + z^{\text{*}} \right)^{4}

\Rightarrow (1 + z)^{4} + \left( 1 + z^{\text{*}} \right)^{4} = \left( 1 + z^{\text{*}} \right)^{4}\left( z^{4} + 1 \right)

Since $z = e^{i\pi/4}$ ,

z^{4} = e^{i\pi} = - 1

so $z^{4} + 1 = 0$ , and

(1 + z)^{4} + \left( 1 + z^{\text{*}} \right)^{4} = \left( 1 + z^{\text{*}} \right)^{4} \times 0 = 0

To check your answer: direct computation with

z = e^{i\pi/4} \approx 0.707 + 0.707i

gives

(1 + z)^{4} + \left( 1 + z^{\text{*}} \right)^{4} \approx 0

✓.

$\therefore$ $(1 + z)^{4} + \left( 1 + z^{\text{*}} \right)^{4} = 0$ .

2021 H2 Maths Paper 1, Question 4

Part (a)

Part (b)(i)

Part (b)(ii)

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