Convergence Properties of the Taylor Series for $\frac{1+z}{1-z}$ 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!
 A: I prefer using geometric series tricks to taking derivatives. Your fraction is
$$
\frac{(1+i)+ (z-i)}{(1 - i) - (z-i)}.
$$
Using $(1+i)(1-i) = 2$, multiply numerator and denominator by $\frac{1+i}{2}$ to get
$$
\frac{\text{stuff}}{1 - \frac{(1+i)(z-i)}{2}}
$$
and so the radius of convergence is $|\frac{2}{1+i}| = \sqrt{2}$ (and you get the explicit formula in the usual way if you're less lazy than I am with the numerator).
A: You shouldn't have that alternating sign term showing up. $$f''(z)=\frac{d}{dz}\bigl[2(1-z)^{-2}\bigr]=2\cdot-2(1-z)^{-3}\cdot\frac{d}{dz}[1-z]=-4(1-z)^{-3}\cdot-1=\frac{4}{(1-z)^3}.$$ Something similar will happen in all other cases, too.
It holds whenever $|z-i|<\sqrt{2}$ because the nearest (in fact, only) singularity is the pole at $z=1$. The distance from $i$ to $1$ is $\sqrt{2}$. The radius of convergence of a Taylor series is the distance from the center to the nearest point of nonanalyticity. You can also, of course, see this by rearranging this as a geometric series. Note that $\frac{2}{1-i}=1+i$, so we can rewrite as $$\frac{1+z}{1-z}=i+(1+i)\sum_{j=1}^\infty\left(\frac{z-i}{1-i}\right)^j=-1+(1+i)\sum_{j=0}^\infty\left(\frac{z-i}{1-i}\right)^j,$$ and the series converges for $$\left|\frac{z-i}{1-i}\right|<1,$$ or equivalently, for $$|z-i|<|1-i|=\sqrt{2}.$$
A: $$
\frac{1+i+u}{1-i-u}=-1+\frac2{1-i}\frac1{1-\frac{u}{1-i}}=-1+\sum\limits_{n=0}^{+\infty}\frac2{(1-i)^{n+1}}u^n,\qquad |u|\lt|1-i|=\sqrt2.
$$
