What is the solution to $\arctan -i$?

Question

I know that $\operatorname{Arctan} z = \frac{1}{2i} \operatorname{Ln} \frac{1+iz}{1-iz}$ but in this case we have $$\frac{1}{2i} \operatorname{Ln}\frac{1+1}{1-1}$$ Is it possible to deal with this formula or what other methods should I try?

Answer 1

Thanks to all! That's what I've got. $\\$ The function $\operatorname{Ln} z$ is considered on the set $\mathbb C \setminus \{ 0 \}$, and the set $\mathbb C $ itself does not include an infinitely distant point. Therefore $\operatorname{Ln}(\infty)$ does not exist. That's why $\operatorname{Arctan} (-i)$ does not exist either.

Answer 2

If $\tan z=-i$ then $\sin^2z=-\cos^2z$, contradicting $\sin^2z+\cos^2z=1$.