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!

Observe that $-1+\sqrt{3}i=2e^{\frac{2\pi i}{3}}$. Then the roots of the equation are
$$
z=\sqrt[4]{2}e^{\frac{\pi i}{6}+\frac{k\pi}{2}},
$$
where $k=0,1,2,3$. Thus $r=\sqrt[4]{2}$. Substitute $k$, then we get $z=\sqrt[4]{2}e^{\frac{\pi i}{6}}$, $z=\sqrt[4]{2}e^{\frac{2\pi i}{3}}$, $z=\sqrt[4]{2}e^{\frac{7\pi i}{6}}$, and $z=\sqrt[4]{2}e^{\frac{5\pi i}{3}}$. There is no root whose argument is $\frac{11\pi}{12}$. What happened? Because your attempt has a flaw: Since $r^4 \cos 4\theta=-1$ and $r^4\sin 4\theta=\sqrt{3}$, we get $\tan 4\theta=-\sqrt{3}$, but it doesn't imply that $4\theta = -\frac{\pi}{3}$. If it is true, then $r^4 \cos4\theta$ cannot be negative!
