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

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.

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• Dividing by zero -> Undefined – bounceback Feb 22 at 18:46
• @Asif Touhid Do you have problem on the phase of this complex number? – pawel Feb 22 at 18:50
• Yes, but can you please compare this (2/0) with (1/0)? – Asif Touhid Feb 22 at 18:51

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: $$\displaystyle \sin\theta=-1\iff \theta=-\frac\pi2$$
• Nothing changes in terms of phase what's change is the modulus and so $r$ (it is 2 or 1 in the second case) – pawel Feb 22 at 19:35