In a text that I am reading: Let $\phi(z)=\frac{1}{2}\log\left(\frac{1+z}{1-z}\right),\omega (z)=z^{n},(n\in\mathbb{N}^{+})$, $$f(z)=\mathrm{Re}\left(\int_{0}^{z}\frac{2\phi'(\zeta)}{1-\omega(\zeta)}d\zeta\right)-\overline{\phi(z)}=\mathrm{Re}\left(\int_{0}^{z}\frac{2}{(1-\zeta^2)(1-\zeta^n)}d\zeta\right)-\overline{\frac{1}{2}\log\left(\frac{1+z}{1-z}\right)}.$$
Then they state that the following: $f(z)$ defined for $n$ odd by
\begin{equation*} \begin{split} f(z)&=\frac{1}{n}\left[\mathrm{Re}\left(\frac{z}{1-z}\right)+\sum_{k=1}^{(n-1)/2}\csc\frac{2k\pi}{n}\arg\left(\frac{1-z e^{-i(2k\pi/n)}}{1-z e^{i(2k\pi/n)}}\right)\right]\\ &+\frac{i}{2}\arg\left(\frac{1+z}{1-z}\right), \end{split} \end{equation*} and for $n$ even by \begin{equation*} \begin{split} f(z)&=\frac{1}{n}\left[\mathrm{Re}\left(\frac{2z}{1-z^2}\right)+\sum_{k=1}^{(n/2)-1}\csc\frac{2k\pi}{n}\arg\left(\frac{1-z e^{-i(2k\pi/n)}}{1-z e^{i(2k\pi/n)}}\right)\right]\\ &+\frac{i}{2}\arg\left(\frac{1+z}{1-z}\right), \end{split} \end{equation*}
But how is it derived? I greatly appreciate any help. Thanks!
Assume for first that $n$ is even, $n=2m$. Then $f(z)=\frac{1}{(1-z^2)(1-z^n)}$ has two double poles in $z=\pm 1$ and simple poles in the other $n$-th roots of unity. Now we compute the partial fraction decomposition of $f(z)$. We have $\frac{1}{1-z^2}=\frac{1}{2}\left(\frac{1}{1-z}+\frac{1}{1+z}\right)$ and, if we set $g(z)=\frac{1}{1-z^n}$,
$$ g(z) = \sum_{k=0}^{n-1}\frac{\operatorname{Res}\left(g(z),z=\exp(2\pi i k/n)\right)}{z-\exp(2\pi i k/n)}=\frac{1}{m(1-z^2)}+\sum_{k=1}^{m-1}\frac{1-\cos(\pi k/n)z}{m\left(1+z^2-2\cos(\pi k/m)\right)}.$$ Now we "just" have to multiply everything by $\frac{1}{2}\left(\frac{1}{1-z}+\frac{1}{1+z}\right)$ and take the primitive.
The technique for odd $n$s is similar.