<s>[INST] Finding the taylor series of $f(z) = 1/(1+z^2)$. I am working on the following exercise:

Find the Taylor expansion of the function $f(z) = \frac{1}{1+z^2}$ about $z = 3i$.

We had the Taylor Series Theorem in the lecture:

Let $D \subset \mathbb C$ be a domain and $f: D \to \mathbb C$ a differentiable function. Then $f$ is analytic in $D$ and for any ball $B_R(z_0) \subset D$ the power series expansion
  \begin{align*}
f(z) = \sum_{n = 0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
\end{align*}
  is valid. Further, if $r \in (0,R)$ then
  \begin{align*}
f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{S_r^+(z_0)} \frac{f(w)}{(w-z_0)^{n+1}}dw,
\end{align*}
  where $S_r^+(z_0) = z_0+re^{it}$ with $0\le t \le 2\pi$.

The solution that I don't understand goes like this: Represent $f$ as partial fractions:
\begin{align*}
f(z) = \frac{1}{1+z^2} = \frac{1}{2\pi i} \left(\frac{1}{z-i} - \frac{1}{z+i}\right).
\end{align*}
Then we compute
\begin{align*}
f^{(n)}(z) = \left(\frac{1}{1+z^2}\right)^{(n)} = \frac{1}{2\pi i}\left(\frac{(-1)^nn!}{(z-i)^{n+1}}-\frac{(-1)^nn!}{(z+i)^{n+1}}\right).
\end{align*}
Then one can plug this in the formula and is done.
I don't understand why one can compute so easily the $n$th derivative.
 [/INST] A naive, pretty useful many times, approach:
$$\frac1{1+z^2}=\frac1{10+6i+(z-3i)^2}=\frac1{10+6i}\frac1{1+\left(\frac{z-3i}{\sqrt{10+6i}}\right)^2}$$
Now, notice that
$$\left|\frac{z-3i}{\sqrt{10+6i}}\right|<1\iff|z-3i|^2<|10+6i|=4\sqrt{10}$$
for which values of $\;z\;$ we can write
$$\frac1{1+\left(\frac{z-3i}{\sqrt{10+6i}}\right)^2}=\sum_{n=0}^\infty(-1)^n\left(\frac{z-3i}{\sqrt{10+6i}}\right)^n$$
</s>