Problem Statement
Solve the equation $z^4 = -8 - 8i\sqrt{3}$, giving your answers in modulus-argument form $r e^{i\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$.
Verified Solution & Marking Scheme
Express the complex number in polar form
Let $w = -8 - 8i\sqrt{3}$.
$$|w| = \sqrt{(-8)^2 + (-8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16$$
$$\tan \alpha = \left|\frac{-8\sqrt{3}}{-8}\right| = \sqrt{3} \implies \alpha = \frac{\pi}{3}$$
Since both real and imaginary parts are negative, $w$ lies in the 3rd quadrant:
$$\operatorname{Arg}(w) = -\left(\pi - \frac{\pi}{3}\right) = -\frac{2\pi}{3}$$
$$w = 16 e^{i(-2\pi/3 + 2k\pi)}$$ for $k = 0, 1, 2, 3$.
Apply fractional De Moivre Theorem for 4th roots
$$z_k = (16)^{1/4} e^{i\left( \frac{-2\pi/3 + 2k\pi}{4} \right)} = 2 e^{i\left( -\frac{\pi}{6} + \frac{k\pi}{2} \right)}$$
Evaluate roots for k = 0, 1, 2, 3 within (-π, π]
For $k = 0$: $z_0 = 2 e^{-i\pi/6}$
For $k = 1$: $z_1 = 2 e^{i(-\pi/6 + \pi/2)} = 2 e^{i\pi/3}$
For $k = 2$: $z_2 = 2 e^{i(-\pi/6 + \pi)} = 2 e^{i 5\pi/6}$
For $k = 3$: $z_3 = 2 e^{i(-\pi/6 + 3\pi/2)} = 2 e^{i 4\pi/3} = 2 e^{-i 2\pi/3}$