Wednesday, 12 April 2017

algebra precalculus - Square root of complex number.

The complex number $z$ is defined by $z=\frac{9\sqrt3+9i}{\sqrt{3}-i}$. Find the two square roots of $z$, giving your answers in the form $re^{i\theta}$, where $r>0$ and $-\pi <\theta\leq\pi$



I got the $z=9e^{\frac{\pi}{3}i}$. So I square root it, it becomes $3e^{\frac{\pi}{6}i}$. But the given answer is $3e^{-\frac{5}{6}\pi i}$. Why?

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