solve: $z^3=\sqrt(3)-i$ 
Solve: $z^3=\sqrt(3)-i$

$r=\sqrt{\sqrt{3}^2+(-1)^2}=\sqrt{4}$
$\theta=tan^{-1}(\frac{-1}{\sqrt{3}})=\frac{-\pi}{6}$
0: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{-\pi}{18})+isin(\frac{-\pi}{18})]=1.24-0.96i$
1: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{11\pi}{18})+isin(\frac{11\pi}{18})]=-0.43-1.18i$
2: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{23\pi}{18})+isin(\frac{23\pi}{18})]=-0.8-0.96i$
Is it right? Wolfram has a much shorter way
 A: $$z^3=\sqrt{3}-i \Longleftrightarrow$$
$$z^3=\left|\sqrt{3}-i\right|e^{\arg\left(\sqrt{3}-i\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{\left(\sqrt{3}\right)^2+1^2}e^{\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{3+1}e^{-\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{4}e^{-\frac{\pi}{6}i} \Longleftrightarrow$$
$$z^3=2e^{-\frac{\pi}{6}i} \Longleftrightarrow$$
$$z=\left(2e^{\left(2\pi k-\frac{\pi}{6}\right)i}\right)^{\frac{1}{3}} \Longleftrightarrow$$
$$z=2^{\frac{1}{3}}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{6}\right)i} \Longleftrightarrow$$
$$z=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{6}\right)i} $$
With $k\in\mathbb{Z}$ and $k:0-2$

So the solutions are:
$$z_0=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 0-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{-\frac{\pi}{18}i}\approx 1.2407-0.2187i$$
$$z_1=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 1-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{\frac{11\pi}{18}i}\approx -0.4309+1.1839i$$
$$z_2=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 2-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{-\frac{13\pi}{18}i}\approx -0.809-0.965i$$
