2023 H2 Maths Paper 1 Q8 — Worked Solution

Part (a)(i)

2 marks

z = 2e^{i2\pi/3}

Explanation: $r = |z| = \sqrt{( - 1)^{2} + \left( \sqrt{3} \right)^{2}} = \sqrt{4} = 2$ . Since $Re(z) = - 1 < 0$ and $Im(z) = \sqrt{3} > 0$ , $z$ lies in the second quadrant, with reference angle $\tan^{- 1}\left( \frac{\sqrt{3}}{1} \right) = \frac{\pi}{3}$ , so $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ .

To check your answer:

2e^{i2\pi/3} = 2\left( \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right) = 2\left( - \frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = - 1 + \sqrt{3}\, i

✓

$\therefore z = 2e^{i2\pi/3}$ .

Part (a)(ii)

3 marks

Approach

with $z$ and $z^{\text{*}}$ both in exponential form, $\frac{z^{n}}{iz^{\text{*}}}$ reduces to a single exponential $2^{n - 1}e^{i\phi}$ ; “purely imaginary” then just needs $\phi \equiv \frac{\pi}{2}\ \left( \text{mod }\pi \right)$ .

Since $z = 2e^{i2\pi/3}$ , $z^{\text{*}} = 2e^{- i2\pi/3}$ , and $i = e^{i\pi/2}$ : $z^{n} = 2^{n}e^{i2n\pi/3}$ , and

iz^{\text{*}}

= e^{i\pi/2} \cdot 2e^{- i2\pi/3}

= 2e^{i(\pi/2 - 2\pi/3)}

= 2e^{- i\pi/6}

\frac{z^{n}}{iz^{\text{*}}}

= \frac{2^{n}e^{i2n\pi/3}}{2e^{- i\pi/6}}

= 2^{n - 1}e^{i(2n\pi/3 + \pi/6)}

For this to be purely imaginary, its argument must satisfy $\frac{2n\pi}{3} + \frac{\pi}{6} \equiv \frac{\pi}{2}\mkern6mu \left( \text{mod }\pi \right)$ :

\frac{2n\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2} + k\pi,\quad k \in \mathbb{Z}

\Rightarrow 4n + 1 = 3 + 6k

\Rightarrow n = \frac{1 + 3k}{2}

For $n$ to be a positive integer, $1 + 3k$ must be a positive even number, so $k$ is odd. The smallest odd $k$ giving a positive $n$ is $k = 1$ , giving $n = 2$ .

To check your answer: at

n = 2

, argument

\text{=}\frac{4\pi}{3} + \frac{\pi}{6} = \frac{3\pi}{2} \equiv \frac{\pi}{2}\ \left( \text{mod }\pi \right)

✓, so

\frac{z^{2}}{iz^{\text{*}}}

is purely imaginary; at

n = 1

, argument

\text{=}\frac{5\pi}{6} ≢ \frac{\pi}{2}\ \left( \text{mod }\pi \right)

, so

n = 1

fails.

$\therefore$ the smallest positive integer is $n = 2$ .

Part (b)

5 marks

Approach

$|w|$ is real, so equating the imaginary part of $2v + |w| = 1$ forces $v$ itself to be real; substituting $v = a$ into the second equation and comparing real and imaginary parts gives $c$ and $d$ in terms of $a$ , leaving one equation in $a$ from (1).

Let $v = a + bi$ and $w = c + di$ , where $a,b,c,d \in \mathbb{R}$ . Since $|w|$ is real, the equation $2v + |w| = 1$ has a real right side, so its imaginary part gives $2b = 0 \Rightarrow b = 0$ : $v = a$ is real. The real part gives

2a + |w| = 1 \Rightarrow 2a + \sqrt{c^{2} + d^{2}} = 1\quad\text{...(1)}

Substituting $v = a$ and $w = c + di$ into $3v - iw = - 3 + 4i$ ,

3a - i(c + di)

= 3a - ci - di^{2}

= (3a + d) - ci

Comparing with $- 3 + 4i$ : $3a + d = - 3\ \text{...(2)}$ and $- c = 4 \Rightarrow c = - 4\ \text{...(3)}$ .

From (2), $d = - 3 - 3a$ . Substituting into (1) with $c = - 4$ ,

2a + \sqrt{16 + ( - 3 - 3a)^{2}} = 1

\Rightarrow \sqrt{16 + (3 + 3a)^{2}} = 1 - 2a

Squaring both sides (valid since $1 - 2a \geq 0$ , i.e. $a \leq 0.5$ ),

16 + (3 + 3a)^{2} = (1 - 2a)^{2}

\Rightarrow 25 + 18a + 9a^{2} = 1 - 4a + 4a^{2}

\Rightarrow 5a^{2} + 22a + 24 = 0

To check your answer: discriminant

\text{=}22^{2} - 4(5)(24) = 4 = 2^{2}

, a perfect square, so

a

has two rational roots ✓

a = \frac{- 22 \pm 2}{10} \Rightarrow a = - 2\ \text{or}\ a = - \frac{12}{5}

Both satisfy $a \leq 0.5$ , so both are valid. Using $d = - 3 - 3a$ : for $a = - 2$ , $d = 3$ ; for $a = - \frac{12}{5}$ , $d = - 3 + \frac{36}{5} = \frac{21}{5}$ .

To check your answer: for

v = - 2,\ w = - 4 + 3i

:

2v + |w| = - 4 + 5 = 1

✓ and

3v - iw = - 6 + 4i + 3 = - 3 + 4i

✓. For

v = - \frac{12}{5},\ w = - 4 + \frac{21}{5}i

:

2v + |w| = - \frac{24}{5} + \frac{29}{5} = 1

✓ and

3v - iw = - \frac{36}{5} + 4i + \frac{21}{5} = - 3 + 4i

✓

$\therefore v = - 2,\ w = - 4 + 3i\quad\text{or}\quad v = - \frac{12}{5},\ w = - 4 + \frac{21}{5}i$ .

2023 H2 Maths Paper 1, Question 8

Part (a)(i)

Part (a)(ii)

Part (b)

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