Answer
$z^2=r^2 [\cos \theta \cos \theta - \sin \theta \sin \theta]+i ( \sin \theta \cos \theta+\cos \theta \sin \theta]$
Work Step by Step
Need to use the FOIL method.
Since, $i^2=-1$
$ \sin \theta$ $ \cos \theta$
Here, $z^2= z \cdot z= [(r \cos \theta )+i (r \sin \theta)] \cdot [(r \cos \theta )+i (r \sin \theta)]$
or, $=r^2 \cos \theta \cos \theta +i r^2 \cos \theta \sin \theta+i r^2 \sin \theta \cos \theta+i^2 r^2 \sin \theta \sin \theta$
or, $=r^2 [\cos \theta \cos \theta +i \cos \theta \sin \theta+i \sin \theta \cos \theta - \sin \theta \sin \theta$
Thus, we have $z^2=r^2 [\cos \theta \cos \theta - \sin \theta \sin \theta]+i ( \sin \theta \cos \theta+\cos \theta \sin \theta]$
