Answer
$z^2=r^2 [\cos 2\theta +i \sin 2\theta]$
Work Step by Step
From the previous part (b), we have
$z^2=r^2 [\cos \theta \cos \theta - \sin \theta \sin \theta]+i ( \sin \theta \cos \theta+\cos \theta \sin \theta]$
Need to use the formulas such as:
$\sin (a+b)= \sin a \cos b+\cos a \sin b$ and $\cos (a+b)= \cos a \cos b- \sin a \sin b$
Thus, we have
$z^2=r^2 [\cos (\theta + \theta) +i \sin (\theta + \theta)]$
Hence, $z^2=r^2 [\cos 2\theta +i \sin 2\theta]$
