I just ran into the following exercise:
Find and state the convergence properties of the Taylor series for the following: $\frac{1+z}{1-z}$ around $z_0=i$.
First of all, let
$f(z)=\frac{1+z}{1-z}.$
Then we have that
$\begin{align} f^{(1)}(z)&=\frac{2}{(1-z)^2},\\ f^{(2)}(z)&=\frac{-4}{(1-z)^3},\\ f^{(3)}(z)&=\frac{12}{(1-z)^4},\cdots \end{align}$
leading us to conclude that
$\frac{d^j}{dz^j}\left[\frac{1+z}{1-z}\right]=2(-1)^{j+1}\left(1-z\right)^{-j-1}j!,$
for $j\geqslant1$. Now, evaluating these at $z=i$ should give the Taylor series
$i+\sum_{j=1}^{\infty}\frac{2}{(1-i)^{j+1}}(z-i)^j.$
I have two questions: I could derive that same series without problems if the $(-1)^{j+1}$ term were not there, but in this case, where does it go? Also, how does the author determine that this holds whenever $|z-i|<\sqrt{2}$? I suppose that this has something to do with the fact that
$\sum_{j=0}^{\infty}c^j$
converges whenever $|c|<1$. Thanks in advance!