Find the principal value of $\big[\frac{e}{2}(-1-\sqrt3i)\big]^{3\pi i}$

Question

I have been asked to find the principal value of: $\big[\frac{e}{2}(-1-\sqrt3i)\big]^{3\pi i}$

The textbook Complex Variables with Applications by A. David Wunsch only provides answers to the odd questions and I have found myself at a loss. All examples I have found have been much simpler in nature. any help would be appreciated.

Answer 1

$1+i\sqrt 3 = 2 (\frac 12 + i \frac {\sqrt {3}}{2}) = 2(\cos \frac {\pi}{3} + i\sin \frac {\pi}{3}) = 2e^{\frac {\pi}{3}i}$

$\left((\frac {e}{2})(2e^{\frac {\pi}{3}i})\right)^{3\pi i}\\ \left(e^{\frac {\pi}{3}i+1}\right)^{3\pi i}\\ e^{(\frac {\pi}{3}i+1)\cdot (3\pi i)}\\ (e^{-\pi^2}e^{3\pi i})\\ e^{3\pi i} = -1\\ -e^{-\pi^2}\\ $

Answer 2

\begin{align} & \left(\frac e2(-1-\sqrt{3}i)\right)^{3\pi i} \\ = & e^{3\pi i}\cdot\left(-\frac12-\frac{\sqrt3}2i\right)^{3\pi i} \\ = & (e^{\pi i})^3\cdot\left(e^{\tfrac{-2\pi i}3}\right)^{3\pi i} \\ = & (-1)^3\cdot e^{\tfrac{-2\pi i}3\cdot3\pi i} \\ = & -e^{2\pi^2} \\ \end{align}