Convert the given complex number into polar form: $-3$.

(N/A) Let the complex number be $z = -3 + 0i$.
We represent it as $z = r(\cos \theta + i \sin \theta)$,where $r \cos \theta = -3$ and $r \sin \theta = 0$.
Squaring and adding both equations:
$r^{2}(\cos^{2} \theta + \sin^{2} \theta) = (-3)^{2} + 0^{2}$
$r^{2} = 9$
Since $r > 0$,we have $r = 3$.
Now,$3 \cos \theta = -3 \Rightarrow \cos \theta = -1$ and $3 \sin \theta = 0 \Rightarrow \sin \theta = 0$.
The angle $\theta$ that satisfies $\cos \theta = -1$ and $\sin \theta = 0$ is $\theta = \pi$.
Thus,the polar form is $3(\cos \pi + i \sin \pi)$.