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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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  • $\begingroup$ Dividing by zero -> Undefined $\endgroup$ – bounceback Feb 22 at 18:46
  • $\begingroup$ @Asif Touhid Do you have problem on the phase of this complex number? $\endgroup$ – pawel Feb 22 at 18:50
  • $\begingroup$ Yes, but can you please compare this (2/0) with (1/0)? $\endgroup$ – Asif Touhid Feb 22 at 18:51
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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: $\displaystyle \sin\theta=-1\iff \theta=-\frac\pi2$

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  • $\begingroup$ Thanks for the clarification. Would you please compare arctan(2/0) with arctan(1/0) please? $\endgroup$ – Asif Touhid Feb 22 at 19:11
  • $\begingroup$ Nothing changes in terms of phase what's change is the modulus and so $r$ (it is 2 or 1 in the second case) $\endgroup$ – pawel Feb 22 at 19:35
  • $\begingroup$ Okay, thanks for your time! $\endgroup$ – Asif Touhid 2 days ago

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