Hello, Janu42!

You never did write it in polar form . . .

Convert $\displaystyle 3\sqrt{3} - 9i$ into polar form. The triangle looks like this: Code:

_
3√3
* - - - - *
* θ |
* |9
r * |
* |
*

$\displaystyle r \:=\:\sqrt{(3\sqrt{3})^2 + 9^2} \;=\;\sqrt{108} \:=\:6\sqrt{3}$

$\displaystyle \cos\theta \:=\:\frac{3\sqrt{3}}{6\sqrt{3}} \:=\:\frac{1}{2} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{3}$

On the graph, we see that the angle is $\displaystyle -\frac{\pi}{3}$

Therefore, the polar form is:

. . $\displaystyle 6\sqrt{3}\bigg[\cos\left(\text{-}\tfrac{\pi}{3}\right) + i\sin\left(\text{-}\tfrac{\pi}{3}\right)\bigg] \;=\;6\sqrt{3}\left(\cos\tfrac{\pi}{3} - i\sin\tfrac{\pi}{3}\right)$