Finding the Taylor series of $f(z)=\frac{1}{z}$ around $z=z_{0}$ I was asked the following (homework) question:

For each (different but constant) $z_{0}\in G:=\{z\in\mathbb{C}:\,
 z\neq0$} find a power series $\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$
  whose sum is equal to $f(z)$ on some subset of $G$. Please specify
  exactly on which subset this last claim holds.
Suggestion: Instead of calculating derivatives of f, try
  using geometric series in some suitable way.

What I did:
Denote $f(z)=\frac{1}{z}$ and note $G\subseteq\mathbb{C}$ is open.
For any $z_{0}\in G$ the maximal $R>0$ s.t $f\in H(D(z_{0},R))$
is clearly $R=|z_{0}|$.
By Taylor theorem we have it that $f$ have a power series in $E:=D(z_{0},R)$
(and this can not be expended beyond this point, as this would imply
that $f$ is holomorphic at $z=0$.
I am able to find $f^{(n)}(z_{0})$ and to solve the exercise this
way: I got that $$f^{(n)}(z)=\frac{(-1)^{n}n!}{z^{n+1}}$$ hence $$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{z_{0}^{n+1}}(z-z_{0})^{n};\, z\in E$$
but I am not able to follow the suggestion, I tried to do a small
manipulation $$f(z)=\frac{1}{z-z_{0}+z_{0}}$$ and tried to work a bit
with that, but I wasn't able to bring it to the form $$\frac{1}{1-\text{expression}}$$
or something similar that uses a known power series .
Can someone please help me out with the solution that involves geometric
series ?
 A: I think you were on the right track:
$$\frac1z=\frac1{z_0+(z-z_0)}=\frac1{z_0}\cdot\frac1{1+\frac{z-z_0}{z_0}}=\frac1z_0\left(1-\frac{z-z_0}{z_0}+\frac{(z-z_0)^2}{z_{0}^2}-\ldots\right)=$$
$$=\frac1z_0-\frac{z-z_0}{z_0^2}+\frac{(z-z_0)^2}{z_0^3}+\ldots+\frac{(-1)^n(z-z_0)^n}{z_0^{n+1}}+\ldots$$
As you can see, this is just what you got but only using the expansion of a geometric series, without the derivatives explicitly kicking in...The above is true whenever
$$\frac{|z-z_0|}{|z_0|}<1\iff |z-z_0|<|z_0|$$
A: Your manipulation was along correct lines. Start with $\frac{1}{z_0+z-z_0}$. If $z_0\ne 0$, we can rewrite this as
$$\frac{1}{z_0}\frac{1}{1-\frac{(-1)(z-z_0)}{z_0}},$$
and it's over, we can recognize the sum of a geometric progression.  This trick comes up fairly often. 
A: $ f(z) $ does not have a Maclaurin series (a Taylor series centered at $ x = 0 $) because its derivative does not exist there. However, it does have a Laurent series there, namely $ \frac{1}{x} $ is the Laurent series as well as the function.
A Taylor series centered at $ x = x_0 $ only exists if the derivatives of $ f $ at $ x_0 $ exist as well. 
