So I am a bit confused about Laurent series and their relation to geometric series. I will give an example but my doubts are more general. The function: $$f(z) = {\frac {{{\rm e}^{z}}}{ \left( z-1 \right) ^{2}}} $$ has a singularity at $z = 1$. I know that the function converges to a Laurent series only in the region $ \left| z-1 \right| > 0$ which is an "infinite circle" centered at 1 (right?). So relating to the geometric series: $ \left( z-1 \right) ^{-2}$ which is obtained by differentiating $ \left( 1-z \right) ^{-1}$. However, this geometric series is only valid for $$ \left| z \right| <1$$ inside the circle with radius $ \left| z \right|=1$ which is not the same as $ \left| z-1 \right| > 0$.
My question is: how can I relate a geometric series with a correct Laurent expansion and find the correct geometric series in "the correct" regions defined by a certain inequality?
Use the fact that\begin{align}\frac{e^z}{(z-1)^2}&=e\frac{e^{z-1}}{(z-1)^2}\\&=e\frac{\displaystyle\sum_{n=0}^\infty\frac{(z-1)^n}{n!}}{(z-1)^2}.\end{align}