Write –9 in polar form.

We have, z = –9

Let $-9=r \cos \theta$ and $0=r \sin \theta$

By squaring and adding, we get

$(-9)^{2}+(0)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$

$\Rightarrow 81=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$\Rightarrow 81=r^{2}$

$\Rightarrow r=9$

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

$\Rightarrow \theta=\pi$

Thus, the required polar form is $9(\cos \pi+i \sin \pi)$