solve the following equations for x exists in c

Question

$a)$ $z^3 = (1-i\sqrt{3})8$

$b)$ $z^2 - (3-2i)z + (1-3i) = 0 $

$c)$ $z^4 + 1 + i\sqrt{3} = 0$

I know for the last two you start by using the quadratic form but I'm not sure what to do for any of them.

Answer 1

HINT (solve a):

$$z^3=8\left(1-i\sqrt{3}\right)\Longleftrightarrow$$ $$z^3=\left|8-8i\sqrt{3}\right|e^{\arg\left(8-8i\sqrt{3}\right)i}\Longleftrightarrow$$ $$z^3=16e^{-\frac{\pi i}{3}}\Longleftrightarrow$$ $$z=\left(16e^{\left(2\pi k-\frac{\pi}{3}\right)i}\right)^{\frac{1}{3}}\Longleftrightarrow$$ $$z=\sqrt[3]{16}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{3}\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-2$