We have the rational function : $$f(x)=\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}\;\;\;,\;\;n\in \mathbb{Z}^{+}$$ It's not hard to prove that : $$\frac{(1+ix)^{n}-1}{(1-ix)^{n}-1}=(-1)^{n}\prod_{k=1}^{n-1}\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\;\;\;,\;\;\xi_{n}^{k}=e^{2\pi i k/n}$$ Now we want to compute $\log f(x)$ for $x>0$. The logarithm of the individual factors can be written as :
$$\log\left(\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\right)=2i\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)+i\pi;\;\;\;\;x>0$$ So, one would expect: $$\log f(x)=-i\pi+2i\pi n+2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)$$ But it looks nothing like what wolframalpha returns. What am i doing wrong here ?
I think that, if you use the standard branch of $\log$ used by Wolfram Alpha, your third identity should read:
$$\log\left(\frac{x+i(\xi_{n}^{k}-1)}{x-i(\xi_{n}^{k}-1)}\right)=2i\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right)-\pi i,\quad x>0.$$
Compare for exampe: arctan and log.
You are using a branch of the logarithm, so in general it is not true that $\log (ab) = \log(a) + \log(b)$. You can check that the difference between the corrected expression $$\log f(x)=-\pi(n-1)i + 2i\sum_{k=1}^{n-1}\tan^{-1}\left(\frac{x}{1-\xi_{n}^{k}}\right),$$ and the original $\log f(x)$ is a multiple of $\pi$.