Finding square roots of complex numbers in polar form

[Milly]

View attachment 4273Helppp for part (ii). I got 3$e^{\frac{1}{6}\theta i}$

 

[Opalg]

Helppp for part (ii). I got 3$e^{\frac{1}{6}\theta i}$ Do you mean $\color{red}{3e^{\frac{1}{6}\pi i}}$?

The two square roots of $re^{\theta i}$ are $\sqrt re^{\frac12\theta i}$ and $\sqrt re^{(\frac12\theta+ \pi) i}$.