What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$?

Question

$\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}= 2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{42}+\frac{2k\pi}{7})$

The problem I have currently is that I have no idea where I made an error.
The correct solution is
$2^\frac{5}{7}(\cos(\frac{5\pi}{6}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{6}+\frac{2k\pi}{7})$.

Answer 1

You just divided twice by $7$.

$$\sqrt[7]{(-\sqrt3-i)^5}=\sqrt[7]{32\left(\cos(5\cdot210^{\circ})+i\sin(5\cdot210^{\circ})\right)}=$$ $$=2^{\frac{5}{7}}\left(\cos\left(150^{\circ}+\frac{360^{\circ}k}{7}\right)+i\sin\left(150^{\circ}+\frac{360^{\circ}k}{7}\right)\right),$$ where $k\in\{0,1,2,3,4,5,6\}$.

Answer 2

$(-\sqrt3-i)^5$ 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

$$\sqrt[7]{(-\sqrt3-i)^5}=2^{5/7}e^{i({5\pi/6}+2\pi k/7)}=2^{5/7}\left(\cos\left({5\pi\over6}+{2\pi k\over7}\right)+i\sin\left({5\pi\over6}+{2\pi k\over7}\right)\right)$$

Answer 3

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 $$ \{ 2^{5/7} e^{-\pi/6 + 2 k \pi/7}, k=0,1,...,6 \}. $$