Find all complex numbers $z$ satisfying the equation $z^{4} = -1+\sqrt{3}i$ Since any complex number can be of polar form. We set that $z = r\cos \theta +ir\sin \theta$. Now by de Moivre's Theorem, we easily see that $$z^{4} = r^{4}\cos4 \theta + ir^{4}\sin4 \theta$$
Since $$z^{4} = -1+\sqrt{3}i$$
We equip accordingly and see that $$r^{4}\cos4 \theta = -1$$
$$r^{4}\sin4 \theta = \sqrt{3}$$
Solving the above 2 equations we have $$\tan 4\theta = -\sqrt{3} \Rightarrow 4\theta = -\dfrac{\pi}{3} \Rightarrow \theta = -\dfrac{\pi}{12}$$
Hence we have $$\text{arg}z = -\dfrac{\pi}{12} + k\pi$$
Furthermore, $$\text{Arg}z = -\dfrac{\pi}{12} \text{ or } \dfrac{11\pi}{12}$$
However, i cannot find out what $r$ is, as when i substitute in to solve, $r$ become a complex number?? Is my answer correct, how can i make my steps better? Thanks!

The absolute value is very simple:
$$|z^4|=|-1+\sqrt3 i|=2$$
Thus,
$$|z|=\sqrt[4]2$$
