Problem Statement
Solve the equation $z^3 = 8i$, giving your answers in the form $r e^{i\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$.
Verified Solution & Marking Scheme
Express 8i in polar/exponential form
Modulus $|8i| = 8$.
Argument: Since $8i$ lies on positive imaginary axis, $\arg(8i) = \frac{\pi}{2}$.
General form with $2k\pi$ periodicity:
$8i = 8 e^{i(\pi/2 + 2k\pi)}, \quad k \in \mathbb{Z}$
Take cube roots using fractional power
$z_k = (8)^{1/3} e^{i\left(\frac{\pi/2 + 2k\pi}{3}\right)} = 2 e^{i\left(\frac{\pi + 4k\pi}{6}\right)}, \quad k = 0, 1, 2$
Evaluate each root for principal argument range (-π, π]
- For $k = 0$: $\theta_0 = \frac{\pi}{6} \implies z_0 = 2 e^{i\pi/6}$
- For $k = 1$: $\theta_1 = \frac{\pi + 4\pi}{6} = \frac{5\pi}{6} \implies z_1 = 2 e^{i 5\pi/6}$
- For $k = -1$ (or $k = 2$ adjusted to $(-\pi, \pi]$): $\theta_2 = \frac{\pi - 4\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2} \implies z_2 = 2 e^{-i\pi/2} = -2i$