What is the polar form of $-6i$?

Question

The module of $-6i$ is $6$ (the square root of $36$), but $ \tan\theta = -\cfrac{6}{0}$, meaning that the polar form $ 6(\cos\theta + i\sin\theta) $ is also indefinite?

Answer 1

You don't need $\tan$, you need angle. Just look for the angle in complex plane. It is $\theta = \frac {3\pi} 2$. So, $\cos \theta = 0$ and $\sin \theta = -1$, $-6i = 6 i \sin \frac {3\pi} 2$. Or just use atan2.

Answer 2

Another way:

You know $r=|-6i|=6.$ Now putting this into polar form we have $$-6i=6(\cos\theta+i\sin\theta)\implies -i=\cos\theta+i\sin\theta\implies \cos\theta=0\ \text{and}\sin\theta=-1\implies\quad \theta=\frac{3\pi}{2}$$