Taylor Series Expansion of $f(z)=\frac{1+z}{1-z}$ centered at $i \in\Bbb{C}$

Question

Am I doing this right?

Attempt at solution:

Note that $\frac{1+z}{1-z}=\frac{1}{1-z}+\frac{z}{1-z}$. Consider that $\frac{d}{dz}(\frac{z}{1-z})=\frac{1}{(1-z)^2}=\frac{d}{dz}(\frac{1}{1-z})$ and the following expansion: \begin{align} \frac{1}{1-z}&=\frac{1}{1-i-z+i}=\frac{1}{1-i-(z-i)}=\frac{1}{(1-i)(1-\frac{z-i}{1-i})}=\frac{1}{1-i}\sum_{k=0}^{\infty}(\frac{z-i}{1-i})^k \end{align} Also, note \begin{align} \int(\frac{z-i}{1-i})^kdz=\frac{1}{(1-i)^k}\int(z-i)^kdz\\ \end{align} \begin{align} \text{Let} \quad u=z-i\quad \text{so} \quad du=dz \quad \text{and}\quad \frac{1}{(1-i)^k}\int(z-i)^kdz=\frac{1}{(1-i)^k}\int u^kdu=\frac{u^{k+1}}{(k+1)(1-i)^k}+C=\frac{(z-i)^{k+1}}{(k+1)(1-i)^k}+C.\quad \text{Thus,}\quad \frac{z}{1-z}=\frac{1}{1-i}\sum_{k=0}^{\infty}\frac{(z-i)^{k+1}}{(k+1)(1-i)^k}\\ \end{align} \begin{align} \text{and we have}\quad \frac{1}{1-z}+\frac{z}{1-z}=\frac{1}{1-i}\Biggl(\sum_{k=0}^{\infty}\biggl(\frac{z-i}{1-i}\biggr)^k + \sum_{k=0}^{\infty}\frac{(z-i)^{k+1}}{(k+1)(1-i)^k}\Biggr) \end{align}

Answer 1

It seems that a straightforward computation, following your first few steps (before the integral appears) gives $$ \frac{1+z}{1-z} = i + \sum_{k \geq 0} \frac{(1+i)^{k+1}}{2^k}(z-i)^k $$ (you may want to replace $(1+i)$ by $2^{1/2}e^{i\pi/4}$ but that doesn't make a huge difference). Are you trying to prove a particular identity?