Assume that I have the following complex function
$$f(Z)=\frac{1}{2\pi i}\left(-0.5a(Z-1)\ln \left( \frac{Z-1}{Z+1} \right) +0.5b(Z+1)\ln \left( \frac{Z-1}{Z+1} \right) \right)$$
where $Z$ is some complex variable, $i$ is $\sqrt{-1}$, and $a$ and $b$ are real constants. It is my goal to find the derivative of $f$ with regards to $Z$, i.e.
$$f'(Z) = \frac{df(Z)}{dZ}=?$$
Since I am not very familiar with complex differentiation, I tried evaluating this in Wolfram Alpha. You can check the result by clicking the link or entering the following query yourself:
derivative of 1/(2*pi*i)*(-a*0.5*(z-1)*log((z-1)/(z+1))+b*0.5*(z+1)*log((z-1)/(z+1))) wrt z
Curiously, the solution Wolfram Alpha returns contains a lot of decimal numbers, as if use was made of some kind of approximation of expansion. Unfortunately, don't have a Wolfram Alpha Pro account, so I cannot check the intermediate steps myself. Symbolab comes up with a similar solution but also skips the critical steps.
Do you know what exactly is going on here? Was there an expansion or approximation used, and if so, where and why?