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:
$$-2^{2 / 3} \sin \left(\frac{\pi}{18}\right)+i 2^{2 / 3} \cos \left(\frac{\pi}{18}\right)$$
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How does one arrive at the given answer?
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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
Best Answer
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.