Laurent series for $ \log (\frac{z+1}{z-1})$

Question

I want to find the Laurent series for $\displaystyle \log (\frac{z+1}{z-1})$

I used the Laurent series of: $\;\displaystyle \ln(1+x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n} x^n $ for $x\in \mathbb{R}, |x|<1$

to conclude: $\;\displaystyle \log (\frac{z+1}{z-1})=\log(1+1/z)-\log(1-1/z)=...=\sum_{k=0}^{\infty} \frac{2}{2k+1}\frac{1}{x^{2k+1}} $

I'm wondering: is this allowed to do this in this way?

Answer 1

Since $\log\left(\frac{z+1}{z-1}\right)$ does not blow up near $z=0$, the Laurent series at $z=0$ would be $$ \pi i+\sum_{k=0}^\infty\frac{2z^{2k+1}}{2k+1} $$ which converges for $|z|\lt1$.