Can anyone show me the process to get the result Undefined (1/0) from Tangent inverse one (-2/0) [ tan−1(-2/0) ]?

Question

I know that $\tan^{-1}(1/0)$ is undefined. But I'm getting a little trouble figuring out this $\tan^{−1}(-2/0)$. The answer of $\tan^{−1}(-2/0)$ will also be Undefined. By the way, I got the problem while solving some complex numbers and finding their argument. The complex number was $0-2i$. Thanks in advance. Asif Touhid.

Answer 1

If we think about a generic complex number as $z=a+ib$ then it is possible to observe that:
$a+ib=r(\cos\theta+i\sin\theta)$ (Argand-Gauss representation), where $r=\sqrt{a^2+b^2}$.

In our case we have $\displaystyle a=0$ (and $b=-2$), and so from the equality above $r\cos\theta=0\implies \theta=\pm\frac{\pi}{2}\implies \sin{\theta}=\pm1$.

In our case we have to look at the negative case since $b<0$ and $r>0$, so: $\displaystyle \sin\theta=-1\iff \theta=-\frac\pi2$