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?

The exponential of an imaginary number is a periodic function with period $2\pi$, so your number $z$ can be represented as $z=9e^{i(\frac{\pi}{3}+2k\pi)}$, and the square roots are:
$$
z^{\frac{1}{2}}=3e^{\frac{1}{2}i(\frac{\pi}{3}+2k\pi)}=3e^{i(\frac{\pi}{6}+k\pi)}
$$
that, for $k=0$ is $3e^{i\frac{\pi}{6}}$ and for $k=1$ is $3e^{i\frac{7\pi}{6}}=3e^{i\frac{-5\pi}{6}}$ (that you can also find for $k=-1$). 
