## Algebra 2 (1st Edition)

$z^2=r^2 [\cos \theta \cos \theta - \sin \theta \sin \theta]+i ( \sin \theta \cos \theta+\cos \theta \sin \theta]$
Need to use the 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 $z^2=r^2 [\cos \theta \cos \theta - \sin \theta \sin \theta]+i ( \sin \theta \cos \theta+\cos \theta \sin \theta]$