The complex number $z$ is defined by $z=\frac{9\sqrt{3}+9i}{\sqrt{3}-i}$. Find the square roots of $z$, giving your answers in the form $re^{i\theta}$.where $r>0$ and $-\pi < \theta \leq \pi$.
I found the $z=9e^{\frac{\pi}{3}i}$. How to find the square roots of $z$?
$\endgroup$ 11 Answer
$\begingroup$Assuming your calculation of $\;z\;$ is correct, observe that
$$z=9e^{\frac{\pi i}3}=9e^{\frac{\pi i}3+2k\pi i}\;,\;\;k\in\Bbb Z\implies z^{1/2}=\left(9e^{\frac{\pi i}3+2k\pi i}\right)^{1/2}=3e^{\frac{\pi i}6\left(6k+1\right)}$$
and this time we only need $\;k=0,1\;$ since the roots repeat themselves, so the roots are
$$\begin{cases}&z_1:=3e^{\frac{\pi i}6}=3\left(\frac{\sqrt3}2+\frac12i\right)=\frac{3\sqrt3}2+\frac32i\\{}\\&z_2=3e^{\frac{7\pi i}6}=3\left(-\frac{\sqrt3}2-\frac12i\right)=-\frac{3\sqrt3}2-\frac32i=-z_1\end{cases}$$
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