## Algebra 2 (1st Edition)

$z^2=r^2 [\cos 2\theta +i \sin 2\theta]$
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, $z^2=r^2 [\cos 2\theta +i \sin 2\theta]$