solve complex equation $x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
What's the way to solve this kind of equation? I think there must be 8 solutions.
I tried to solve the following two equations
$a^6 = 1+i$
$b^6 = \sqrt{3} - i$
I tried the following for the first equation
$-1 = i^2 = (a^6 - 1)^2 = a^{12} - 2a^6 +1$
Then I would need to solve this:
$a^{12} - 2a^6 +2 = 0$
Is this the right approach?

HINT:
$$x^8=\frac{1+i}{\sqrt{3}-i}=\frac{1+i}{\sqrt{3}-i}.\frac{\sqrt{3}+i}{\sqrt{3}+i}=\frac{\sqrt{3}-1}{4}+i\frac{1-\sqrt{3}}{4}$$ 
Then use polar coordinates, $x^8=r(cos\theta +i\sin\theta)$ to obtain all the 8 roots.
Think of the identity $(r(cos\theta +i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$
(This is De-Moivre's Theorem)
