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 $2^\frac{5}{7}(\cos(\frac{5\pi}{6}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{6}+\frac{2k\pi}{7})$.

$(-\sqrt3-i)^5$ is a specific complex number, but it can be written in multiple ways, making use of the fact that $e^{i2\pi k}=1$ for any $k\in\mathbb{Z}$: Since
$$-\sqrt3-i=-2\left(\sqrt3+i\over2\right)=-2\left(\cos\left(\pi\over6\right)+i\sin\left(\pi\over6\right)\right)=-2e^{i\pi/6}$$
we have
$$(-\sqrt3-i)^5=-2^5e^{i5\pi/6}=2^5e^{i11\pi/6}=2^5e^{i23\pi/6}=2^5e^{i35\pi/6}=2^5e^{i47\pi/6}=\cdots$$
Now among the numbers $11$, $23$, $35$, $47$, etc., we see one that's nicely divisible by $7$, so writing
$$(-\sqrt3-i)^5=2^5e^{i(35\pi/6+2\pi k)}$$
we get
$$\sqrt[7]{(-\sqrt3-i)^5}=2^{5/7}e^{i({5\pi/6}+2\pi k/7)}=2^{5/7}\left(\cos\left({5\pi\over6}+{2\pi k\over7}\right)+i\sin\left({5\pi\over6}+{2\pi k\over7}\right)\right)$$
