2022 H2 Maths Paper 2 Q3 — Worked Solution

Part (a)

3 marks

Since $z_{1} = 3 - i\sqrt{3}$ ,

\left| z_{1} \right| = \sqrt{3^{2} + \left( \sqrt{3} \right)^{2}} = \sqrt{12} = 2\sqrt{3}

$z_{1}$ has positive real part and negative imaginary part, so it lies in the fourth quadrant, giving

\arg z_{1} = - \tan^{- 1}\left( \frac{\sqrt{3}}{3} \right) = - \tan^{- 1}\left( \frac{1}{\sqrt{3}} \right) = - \frac{\pi}{6}

Also $z_{2} = \frac{1}{2}e^{i\frac{2\pi}{5}}$ , so $\left| z_{2} \right| = \frac{1}{2}$ and $\arg z_{2} = \frac{2\pi}{5}$ . Since $z_{3} = z_{1} \times z_{2}$ ,

\left| z_{3} \right| = \left| z_{1} \right|\left| z_{2} \right| = 2\sqrt{3} \times \frac{1}{2} = \sqrt{3}

\arg z_{3} = \arg z_{1} + \arg z_{2} = - \frac{\pi}{6} + \frac{2\pi}{5} = - \frac{5\pi}{30} + \frac{12\pi}{30} = \frac{7\pi}{30}

To check your answer: computing

z_{1}z_{2}

directly on the GC gives

\left| z_{3} \right| \approx 1.732

, matching

\sqrt{3} \approx 1.732

, and

\arg z_{3} \approx 0.733

rad, matching

\frac{7\pi}{30} \approx 0.733

rad ✓

$\therefore$ $\left| z_{3} \right| = \sqrt{3}$ and $\arg z_{3} = \frac{7\pi}{30}$ .

Part (b)

2 marks

Part (c)

4 marks

Approach

$z_{3}^{\, n}$ is purely imaginary exactly when $\arg\left( z_{3}^{\, n} \right) = n\arg z_{3}$ is an odd multiple of $\frac{\pi}{2}$ . Since $\arg z_{3}$ is a fraction of $\pi$ with denominator $30$ , this condition becomes a linear congruence in $n$ modulo $30$ , which we solve using the modular inverse of $7$ .

$z_{3}^{\, n}$ is purely imaginary when

n \times \frac{7\pi}{30} = \frac{\pi}{2} + k\pi\quad\text{for some integer}\ k

Dividing by $\pi$ and clearing denominators,

\frac{7n}{30} = \frac{1}{2} + k

\Rightarrow 7n = 15 + 30k

\Rightarrow 7n \equiv 15\mkern6mu \left( \text{mod }30 \right)

Since $\gcd(7,30) = 1$ , $7$ has an inverse modulo $30$ : as $7 \times 13 = 91 = 90 + 1 \equiv 1\mkern6mu \left( \text{mod }30 \right)$ , the inverse is $13$ . Multiplying both sides by $13$ ,

n \equiv 15 \times 13\mkern6mu \left( \text{mod }30 \right)

\Rightarrow n \equiv 195\mkern6mu \left( \text{mod }30 \right)

\Rightarrow n \equiv 15\mkern6mu \left( \text{mod }30 \right)

n = 15

To check your answer:

7 \times 15 = 105 = 3 \times 30 + 15

, so

105 \equiv 15\mkern6mu \left( \text{mod }30 \right)

✓, confirming

n = 15

satisfies

7n \equiv 15\mkern6mu \left( \text{mod }30 \right)

For this value of $n$ ,

\left| z_{3}^{15} \right| = \left| z_{3} \right|^{15} = \left( \sqrt{3} \right)^{15} = 3^{15/2} = 3^{7}\sqrt{3} = 2187\sqrt{3}

\arg\left( z_{3}^{15} \right) = 15 \times \frac{7\pi}{30} = \frac{7\pi}{2}

Reducing to the principal range $( - \pi,\pi\rbrack$ by subtracting $2\pi$ twice,

\frac{7\pi}{2} - 2\pi = \frac{3\pi}{2},\quad\frac{3\pi}{2} - 2\pi = - \frac{\pi}{2}

To check your answer: computing

z_{3}^{15}

directly on the GC gives modulus

\text{≈}3788.0

, matching

2187\sqrt{3} \approx 3788.0

, and argument

\text{≈} - 1.5708

rad, matching

- \frac{\pi}{2} \approx - 1.5708

rad — a purely negative-imaginary number, confirming

z_{3}^{15}

is purely imaginary ✓

$\therefore$ $n = 15$ , $\left| z_{3}^{15} \right| = 2187\sqrt{3}$ and $\arg\left( z_{3}^{15} \right) = - \frac{\pi}{2}$ .

2022 H2 Maths Paper 2, Question 3

Part (a)

Part (b)

Part (c)

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