$x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
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?
Any equation of the form $z^n=\alpha$, where $\alpha$ is a fixed complex number can be solved by switching $\alpha$ to polar form
$$\alpha= r (\cos(\theta)+i\sin(\theta) ) \,.$$
Now, can you find the polar forms of $z$?
For second equation, since the equation only has powers of $a^6$, a substitution $t=a^6$ should work ;)