Problem Statement
Let $z = 1 + i\sqrt{3}$.
(a) Express $z$ in the form $r e^{i\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$. [2]
(b) Hence, find the smallest positive integer $n$ such that $z^n$ is a real number. [3]
Verified Solution & Marking Scheme
(a) Find modulus r and argument θ
Modulus:
$$r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$
Argument:
Since $x = 1 > 0$ and $y = \sqrt{3} > 0$, $z$ lies in the first quadrant:
$$\theta = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$$
Therefore: $$z = 2 e^{i\pi/3}$$
(b) Apply de Moivre's theorem to zⁿ
$$z^n = (2 e^{i\pi/3})^n = 2^n e^{i n\pi/3} = 2^n \left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right)$$
For $z^n$ to be purely real, the imaginary part must equal zero:
$$\sin\left(\frac{n\pi}{3}\right) = 0$$
$$\frac{n\pi}{3} = k\pi, \quad k \in \mathbb{Z} \implies n = 3k$$
For the smallest positive integer $n$ ($k = 1$):
$$n = 3$$
When $n = 3$, $z^3 = 2^3 e^{i\pi} = 8(-1) = -8 \in \mathbb{R}$.