<s>[INST] Taylor Series of $(\frac{z}{1 - z})^2$ around $z = i$ I am having some trouble trying to find the Taylor series of $f(z) = \left(\frac{z}{1 - z}\right)^2$ around $z = i$. I have started a little bit, but unsure how to complete. Below is my attempt:
Since $f(z) = \left(\frac{z}{1 - z}\right)^2$, we are able to rewrite this as:
\begin{equation*}
f(z) = \left(\frac{1}{1 - z} - 1\right)^2 = \dfrac{1}{(1 - z)^2} - \dfrac{2}{1 - z} + 1
\end{equation*}
Observe that $2\frac{1}{1 - z}$ is the Taylor series:
\begin{equation*}
2\frac{1}{1 - z} = 2\sum_{n = 0}^{\infty} \frac{(z - i)^n}{(1 - i)^{n + 1}}
\end{equation*}
around $z = i$
Also observe that the Taylor series of $\frac{1}{(1 - z)^2}$ is the Taylor series:
\begin{equation*}
\frac{1}{(1 - z)^2} = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}}
\end{equation*}
around $z = i$
Then we have that:
\begin{equation*}
f(z) = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1
\end{equation*}
I am not sure how else to proceed from here, and would like some assistance. Thank you.
 [/INST] Observe that for
\begin{equation*}
f(z) = \sum_{n = 0}^{\infty} \dfrac{(z - i)^{n + 1}}{(1 - i)^{n + 2}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1
\end{equation*}
you may reshift the indices of the first summation so that
\begin{equation*}
f(z) = \sum_{n = 1}^{\infty} \dfrac{(z - i)^{n}}{(1 - i)^{n + 1}} - \sum_{n = 0}^{\infty} \dfrac{2(z - i)^n}{(1 - i)^{n + 1}} + 1.
\end{equation*}
From here it is just trivial simplifications, namely taking out one term in the second summation and then combining them again.
$$ f(z) = 1 - \frac{2}{1-i} - \sum_{n=1}^\infty \frac{(z-i)^n}{(1-i)^{n+1}}\\
\implies 
f(z) = -\frac{1+i}{1-i} - \sum_{n=1}^\infty \frac{(z-i)^n}{(1-i)^{n+1}} \\
\implies f(z) = -i - \sum_{n=1}^\infty \frac{(z-i)^n}{(1-i)^{n+1}}$$
If you wanted to you could shift it once more to start at $ n  = 0$ as that's usually the convention.
</s>