Arctan in complex logarithm form

As @gammatester mentioned, you forgot an integration constant. But what value should this integration constant be? We do know that we want $\arctan{(0)} = 0$. Then we would have

$$\frac{\log{(-i)} - \log{(i)}}{2 i} + C = 0$$

What is $\log{(-i)}$? Is $-i = e^{-i \pi/2}$, or $e^{i 3 \pi/2}$, or something else? Regardless, the question is, is the difference between the logs then $i \pi$ or $-i \pi$?

Here we note that the principal value of the arctan is defined with respect to a branch cut $-i z \in (-\infty,-1] \cup [1,\infty) $. Thus, we would take $-i = e^{-i \pi/2}$ and the difference between the logs to be $-i \pi$; thus, $C=\pi/2$. Thus

$$\arctan{z} = \frac1{2 i} \left [\log{(z-i)} - \log{(z+i)} - i \pi \right ] = \frac{i}{2} \log{\left ( \frac{i+z}{i-z}\right )}$$

$$\text{arctanh}(iz)=\sum_{i=0}\frac{(iz)^{2k+1}}{(2k+1)!}$$

but $$i^{2k+1}=i^{2k}i=\begin{cases}-i&\text{if }k\text{ is odd}\\i&\text{if }k\text{ is even},\\\end{cases}$$ then, $$\sum_{i=0}^\infty \frac{(iz)^{2k+1}}{(2k+1)!}=i\sum_{i=0}^\infty \frac{(-1)^kz^{2k+1}}{(2k+1)!}=i\arctan(z)$$ and thus $$\arctan(z)=\frac{\text{arctanh}(iz)}{i}.$$

But $\text{arctanh}(z)=\frac{1}{2}\ln\left(\frac{1+z}{1-z}\right)$ and so, we conclude that, $$\arctan(z)=\frac{1}{2i}\ln\left(\frac{1+iz}{1-iz}\right)=\frac{\ln(1+iz)-\ln(1-iz)}{2i}.$$

@gammatester: Yes, you are right! I missed that one! So using

(1) $e^{i\frac{\pi}{2}} = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = i$

in my formula

$\frac{\log(z-i) - \log(z+i)}{2i} + C ~~~~~~~~~$ for $z=0$

I get $C=\frac{\pi}{2}$.

Then using (1) again I can rewrite this as

$\frac{\log(z-i) - \log(z+i)}{2i} - i \cdot \log(i) =\frac{\log(z-i) +\log(i) - (\log(z+i)-\log(i))}{2i}$

By applying logarithm laws I then get the same result as @idm !

$\frac{\log(z-i) +\log(i) - (\log(z+i)-\log(i))}{2i} = \frac{\log(1+iz) - \log(1-iz)}{2i} = \arctan(z)$

Many Thanks to both of you! :)