Convert the given complex number in polar form: –3

$-3$

Let $r \cos \theta=-3$ and $r \sin \theta=0$

On squaring and adding, we obtain

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(-3)^{2}$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=9$

$\Rightarrow r^{2}=9$

$\Rightarrow r=\sqrt{9}=3 \quad[$ Conventionally, $r>0]$

$\therefore 3 \cos \theta=-3$ and $3 \sin \theta=0$

$\Rightarrow \cos \theta=-1$ and $\sin \theta=0$

$\therefore \theta=\pi$

$\therefore-3=r \cos \theta+i r \sin \theta=3 \cos \pi+\hat{B} \sin \pi=3(\cos \pi+i \sin \pi)$

This is the required polar form.