Laurent series of $e^z$

Question

find the Laurent series centered at $z=1$ $$ f(z)=\frac{e^z}{(z-1)^2} $$ I thought that the denominator part is safe by our center and the expansion is just about the exponential which is a Taylor series but that doesn't match the calculator solution. Any help is appreciated.

Solution:

so we are good at $(z-1)^2$, then we just need to do Taylor expansion for $e^z$ at $z=1$.( that's the center for our Laurent series), which would be $$ f(z) = \frac{1}{(z-1)^2}\sum \frac{e}{k!}(z-1)^k $$ thanks for everyone making the hint!

Answer 1

First notice that $z=1$ is a singularity of $\displaystyle f(z)=\frac{e^{z}}{(z-1)^{2}}$. Setting the change of variables $u=z-1$, we have $u+1=z$. Thus, rewriting all depending of $u$, we have \begin{align*}\frac{e^{z}}{(z-1)^{2}}&=\frac{e^{u+1}}{u^{2}},\\&=\frac{e}{u^{2}}\cdot e^{u},\\&=\frac{e}{u^{2}}\left(1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\cdots\right),\quad |u|<+\infty \\ &=\frac{e}{u^{2}}+\frac{e}{u}+\frac{e}{2!}+\frac{e}{3!}u+\cdots,\\&=\frac{e}{(z-1)^{2}}+\frac{e}{z-1}+\frac{e}{2!}+\frac{e}{3!}(z-1)+\cdots\end{align*} with $z=1$ a pole of order $2$ and the series converges for all $z\not=1$.

Therefore the Laurent Series around of the singularity $z=1$ is given by \begin{align*}\frac{e^{z}}{(z-1)^{2}}&=\frac{e}{(z-1)^{2}}+\frac{e}{z-1}+\frac{e}{2!}+\frac{e}{3!}(z-1)+\cdots,\\&=\sum_{n=-2}^{+\infty}\frac{e}{(2+n)!}(z-1)^{n}\end{align*} with convergence for all complex $z$ except $z=1$.