\(Solve\ z^3 = 4\sqrt{3}- 4i\)

\(Express\ your\ answers\ in\ exponential\ form,\ given \ -\pi \leqslant \theta \leqslant \pi\)

\begin{align} & A\ polynomial\ of\ degree\ three\ has\ three\ solutions: let\ these\ be\ \omega_{1},\ \omega_{2},\ \omega_{3} \\ \\ & Begin\ by\ expressing\ our\ equation\ in\ modulus-argument\ form\\ & z^3 = 8\left(cos\left(-\frac{\pi}{6}\right) +isin\left(-\frac{\pi}{6}\right)\right) \\ & \quad = 8\left(cos\left(-\frac{\pi}{6}+2k\pi \right)+ isin \left(-\frac{\pi}{6}+2k\pi \right)\right) \\ \\ & Then\ taking\ the\ cube\ root\ of\ both\ sides,\ our\ solutions\ will\ be\ given\ by \\ & \omega_{k+1}=(z^3)^{\frac{1}{3}} = \left[8\left(cos\left(-\frac{\pi}{6}+2k\pi \right)+ isin \left(-\frac{\pi}{6}+2k\pi \right)\right)\right]^\frac{1}{3}\ where\ k=0,\ 1,\ 2 \\ \\ & Applying\ deMoivre's\ Theorem\ provides\ our\ solutions \\ & for\ k=0, \quad \omega_{1}= 2 \left(cos\left(\frac{-\frac{\pi}{6}+0\pi}{3}\right) +isin\left(\frac{-\frac{\pi}{6}+0\pi}{3}\right) \right) = 2e^{-\frac{\pi}{18} } \\ & for\ k=1,\quad \omega_{2}= 2\left(cos\left(\frac{-\frac{\pi}{6}+2\pi}{3}\right) +isin\left(\frac{-\frac{\pi}{6}+2\pi}{3}\right) \right) = 2e^{\frac{11\pi}{18} } \\ & for\ k=2,\quad \omega_{3} = 2\left(cos\left(\frac{-\frac{\pi}{6}+4\pi}{3}\right) +isin\left(\frac{-\frac{\pi}{6}+4\pi}{3}\right) \right)= 2e^ {\frac{23\pi}{18} } = 2e^ {-\frac{13\pi}{18} }\\ \end{align}