How to solve a complex equation $w^4 = \sqrt{3} -i$

Question

$z = \sqrt{3} -i$

How do I solve a complex equation $w^4 = \sqrt{3} -i$ I know that I first have to rewrite z to polar format which I have done as $z = 2(cos (-π/6) + sin (-π/6))$ but I do not know how to solve $w^4 = \sqrt{3} -i$.

Answer 1

Set $w=r(cosv+isinv)$, then $w^4=r^4(cos4v+isin4v)$ by De Moivre's formula. If you compare this expression with a conversion of $z=\sqrt3 -i$ to polar form, you get:
$w^4=r^4(cos4v+isin4v)=2(cos(11\pi/6)+isin(11\pi/6))$.
Can you do the rest from here?

Answer 2

Your conversion is wrong. Notice that the vertical component is negative but the horizontal is positive, therefore you are in the fourth quadrant. This means your angle is between $3 \pi /2$ and $2 \pi$.

Hint: Write $w = r e^{i \theta}$ and correct the conversion as I have explained above.