Answer
The complex square roots of the complex numbers are,
$3\left[ \cos {{195}^{{}^\circ }}+i\sin {{195}^{{}^\circ }} \right]\ \ \text{ and }\ 3\left[ \cos {{15}^{{}^\circ }}+i\sin {{15}^{{}^\circ }} \right]\ \ \ $.
Work Step by Step
Here
$z=9\left( \cos {{30}^{{}^\circ }}+i\sin {{30}^{{}^\circ }} \right)$
Therefore
$\begin{align}
& {{z}_{1}}=\sqrt[2]{9}\left[ \cos \left( \frac{{{30}^{{}^\circ }}+{{360}^{{}^\circ }}}{2} \right)+i\sin \left( \frac{{{30}^{{}^\circ }}+{{360}^{{}^\circ }}}{2} \right) \right]\ \ \ \\
& {{z}_{1}}=3\left[ \cos \left( \frac{{{390}^{{}^\circ }}}{2} \right)+i\sin \left( \frac{{{390}^{{}^\circ }}}{2} \right) \right]\ \ \ \\
\end{align}$ \
Similarly, for the other root,
$\begin{align}
& {{z}_{0}}=\sqrt[2]{9}\left[ \cos \left( \frac{{{30}^{{}^\circ }}+{{360}^{{}^\circ }}\times 0}{2} \right)+i\sin \left( \frac{{{30}^{{}^\circ }}+{{360}^{{}^\circ }}\times 0}{2} \right) \right]\ \ \ \\
& {{z}_{0}}=3\left[ \cos {{15}^{{}^\circ }}+i\sin {{15}^{{}^\circ }} \right]\ \ \ \\
\end{align}$
The square roots of the complex number in the polar form is $3\left[ \cos \left( \frac{{{390}^{{}^\circ }}}{2} \right)+i\sin \left( \frac{{{390}^{{}^\circ }}}{2} \right) \right]\ $ and $3\left[ \cos {{15}^{{}^\circ }}+i\sin {{15}^{{}^\circ }} \right]\ \ \ $
