<s>[INST] Using De Moivre's law to compute $(-\sqrt3+i)^{2/3}$ Question: 

If $z=-\sqrt{3}+i$, then $z^{2 / 3} = ?$

My work (which is wrong but I am not sure why): 
We can write $z = r(\cos\theta + i\sin\theta)$
$$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$$
$$\theta = \tan^{-1} \left(\frac{-1}{\sqrt{3}}\right) = \frac{5\pi}{6} \quad (*)$$
Therefore 
$$z = 2\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) \right) $$
so (applying De Moivre's)
$$ z^{\frac{2}{3}} = 2^{\frac{2}{3}} \left(\cos\left(\frac{5\pi}{9}\right) + i\sin\left(\frac{5\pi}{9}\right) \right)$$
However, the answer is apparently: 
$$-2^{2 / 3} \sin \left(\frac{\pi}{18}\right)+i 2^{2 / 3} \cos \left(\frac{\pi}{18}\right)$$


*

*How does one arrive at the given answer?

*Forgive my remembrance of high school trigonometry. In $(*)$ I understand that we can take the negative out of $\tan^{-1}$ resulting in
$$\theta=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6} = \frac{11\pi}{6}$$
which results in a different $\theta$ than I had. My intuition for why $\theta =\frac{5 \pi}{6}$ is because $\frac{-1}{2} \div \frac{\sqrt{3}}{2}$ as $\sin \frac{5 \pi}{6} = -\frac{1}{2}$ and $\cos \frac{5 \pi}{6} = \frac{\sqrt{3}}{2}$. But I am not sure if this is the correct reasoning
 [/INST] You forgot $i$ when writing $z = r(\cos \theta + i \sin \theta)$. Arctangent was computed correctly, so
$$
z = 2\left(\cos(\frac{5}{6}\pi) + i \sin(\frac{5}{6}\pi)\right),
$$
$$
z^{2/3} = 2^{2/3}\left(\cos(\frac{5}{9}\pi) + i \sin(\frac{5}{9}\pi)\right).
$$
Now some trigonometry:
$$
\cos(\frac{5}{9}\pi) = \cos(\frac{1}{2}\pi + \frac{1}{18}\pi) = -\sin(\frac{1}{18}\pi).
$$
Similarly,
$$
\sin(\frac{5}{9}\pi) = \cos(\frac{1}{18}\pi),
$$
and you arrive to the answer.
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