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.


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!

  • $\begingroup$ What is your answer? Can you write it down. $\endgroup$ May 16, 2022 at 2:47
  • $\begingroup$ updated. thought it naively but the calculator gave me something else. $\endgroup$
    – 79999
    May 16, 2022 at 2:48
  • $\begingroup$ Can you also post what did your calculator show? $\endgroup$ May 16, 2022 at 2:50
  • 2
    $\begingroup$ Why does your title say one thing and your body another? $\endgroup$ May 16, 2022 at 2:59
  • 1
    $\begingroup$ @79999 You should indicate the limits of the summation. $\endgroup$
    – Gary
    May 16, 2022 at 3:10

1 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 $z=1$.

  • $\begingroup$ is that $(z-1)^n$ instead not 2? $\endgroup$
    – 79999
    May 18, 2022 at 3:41
  • $\begingroup$ Yes, it was a typo. I fixed it, thank you $\endgroup$
    – A. P.
    May 18, 2022 at 11:02

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