# Laurent series of $e^z$

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!

• What is your answer? Can you write it down. May 16, 2022 at 2:47
• updated. thought it naively but the calculator gave me something else. May 16, 2022 at 2:48
• Can you also post what did your calculator show? May 16, 2022 at 2:50
• Why does your title say one thing and your body another? May 16, 2022 at 2:59
• @79999 You should indicate the limits of the summation.
– Gary
May 16, 2022 at 3:10

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 $$z=1$$.
• is that $(z-1)^n$ instead not 2? May 18, 2022 at 3:41