2019 H2 Maths Paper 1 Q9 — Worked Solution

Part (i)(a)

2 marks

Since $|w| = 1$ , we have $w^{- 1} = \overline{w}$ . By De Moivre’s theorem applied to $w = \cos\theta + i\sin\theta$ ,

\frac{1}{w} = w^{- 1} = \cos( - \theta) + i\sin( - \theta)

= \cos\theta - i\sin\theta.

Adding this to $w$ ,

w + \frac{1}{w} = \left( \cos\theta + i\sin\theta \right) + \left( \cos\theta - i\sin\theta \right)

= 2\cos\theta,

which is real since $\theta$ is real. $\ \$ (shown)

Part (i)(b)

4 marks

Approach

$w - 1$ and $w + 1$ each split into a real and an imaginary part in $\theta$ . Rewriting each part with the half-angle identities $\cos\theta - 1 = - 2\sin^{2}\frac{1}{2}\theta$ , $\cos\theta + 1 = 2\cos^{2}\frac{1}{2}\theta$ and $\sin\theta = 2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta$ turns both the numerator and the denominator into a common factor of $\left( \cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta \right)$ , which then cancels.

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

= - 2\sin^{2}\frac{1}{2}\theta + 2i\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta

= 2\sin\frac{1}{2}\theta\left( - \sin\frac{1}{2}\theta + i\cos\frac{1}{2}\theta \right)

= 2i\sin\frac{1}{2}\theta\left( \cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta \right),

since $i\left( \cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta \right) = i\cos\frac{1}{2}\theta - \sin\frac{1}{2}\theta$ .

Similarly,

w + 1 = \left( \cos\theta + 1 \right) + i\sin\theta

= 2\cos^{2}\frac{1}{2}\theta + 2i\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta

= 2\cos\frac{1}{2}\theta\left( \cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta \right).

Dividing,

\frac{w - 1}{w + 1} = \frac{2i\sin\frac{1}{2}\theta\left( \cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta \right)}{2\cos\frac{1}{2}\theta\left( \cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta \right)}

= i\frac{\sin\frac{1}{2}\theta}{\cos\frac{1}{2}\theta}

= i\tan\frac{1}{2}\theta.\ \ \text{(shown)}

$\therefore$ $k = i$ .

To check your answer: at

\theta = \frac{\pi}{2}

,

w = i

, so

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

, and

i\tan\frac{\pi}{4} = i(1) = i

. ✓

Part (ii)

5 marks

Approach

With $|z| = 1$ we have $z\overline{z} = 1$ , so $\overline{z} = \frac{1}{z}$ . This lets us compute $\left| \frac{z - 3i}{1 + 3iz} \right|^{2}$ by multiplying numerator and denominator by their own conjugates and simplifying using $z\overline{z} = 1$ , rather than substituting $z = \cos\phi + i\sin\phi$ and expanding trigonometric terms.

Since $|z| = 1$ , $z\overline{z} = 1$ . Consider the square of the modulus of the numerator:

|z - 3i|^{2} = (z - 3i)\left( \overline{z} + 3i \right)

= z\overline{z} + 3iz - 3i\overline{z} - 9i^{2}

= 1 + 9 + 3i\left( z - \overline{z} \right)

= 10 + 3i\left( z - \overline{z} \right).

Now the square of the modulus of the denominator:

|1 + 3iz|^{2} = (1 + 3iz)\left( 1 - 3i\overline{z} \right)

= 1 - 3i\overline{z} + 3iz - 9i^{2}z\overline{z}

= 1 + 9(1) + 3i\left( z - \overline{z} \right)

= 10 + 3i\left( z - \overline{z} \right).

The two are equal, so

\left| \frac{z - 3i}{1 + 3iz} \right|^{2} = \frac{|z - 3i|^{2}}{|1 + 3iz|^{2}} = 1.

$\therefore$ the modulus of $\frac{z - 3i}{1 + 3iz}$ is $1$ .

2019 H2 Maths Paper 1, Question 9

Part (i)(a)

Part (i)(b)

Part (ii)

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