<s>[INST] What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{42}+\frac{2k\pi}{7})$
The problem I have currently is that I have no idea where I made an error.  The correct solution is $2^\frac{5}{7}(\cos(\frac{5\pi}{6}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{6}+\frac{2k\pi}{7})$.
 [/INST] What you want is the set of solutions to the equation
$$
x^7 = (-\sqrt3 -i)^5.
$$
Now, taking the polar form of $-\sqrt3 -i$ you get
$$
(-\sqrt3 -i)^5 = \left(2\, e^{-5\pi/6}\right)^5 
= 2^5 e^{-25\pi/6}
= 2^5 e^{-\pi/6}
$$
therefore the solutions are 
$$
\{
2^{5/7} e^{-\pi/6 + 2 k \pi/7}, k=0,1,...,6
\}.
$$
</s>