Problem Statement
Consider the complex number $z = 1 + i\sqrt{3}$.
**(a)** [2 marks] Express $z$ in modulus-argument form $r e^{i\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$.
**(b)** [2 marks] Using De Moivre's theorem, find the smallest positive integer $n$ such that $z^n$ is a real number.
**(c)** [3 marks] Find the three cube roots of $z$ in polar form.
Verified Solution & Marking Scheme
Part (a) [A1]
$r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$
Part (a) [A1]
$\theta = \arg(z) = \arctan(\sqrt{3}/1) = \frac{\pi}{3}$. So $z = 2 e^{i\pi/3}$
Part (b) [M1]
$z^n = 2^n e^{i n \pi / 3} = 2^n \left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right)$
Part (b) [A1]
For $z^n \in \mathbb{R}$, $\sin\frac{n\pi}{3} = 0 \implies \frac{n\pi}{3} = k\pi \implies n = 3k$. Smallest $n = 3$.
Part (c) [M1]
$w_k = 2^{1/3} e^{i(\frac{\pi/3 + 2k\pi}{3})} = \sqrt[3]{2} e^{i(\frac{\pi + 6k\pi}{9})}$ for $k = 0, 1, 2$
Part (c) [A1]
$w_0 = \sqrt[3]{2} e^{i\pi/9}$, $w_1 = \sqrt[3]{2} e^{i 7\pi/9}$
Part (c) [A1]
$w_2 = \sqrt[3]{2} e^{i 13\pi/9} = \sqrt[3]{2} e^{-i 5\pi/9}$