I'm trying to calculate the residue of the following function:


using Laurent Expansion. However, there seem to be some steps I am missing and would highly appreciate any help on that topic. I've noticed that there are two isolated singularities (poles) at $z=1$ and $z=-1$ and I'm able to calculate the residue using the residue formula at the poles.However, I can't seem to get the Laurent Expansion part to make sense.

For instance, the residue at $z=1$ is $e/2$.

Expanding $e^z$ at $z=1$:

$e^z = e(1+(z-1)+(z-1)^2/2!+(z-1)^3/3!+...$

Now comes the part that doesn't fit: based on the following course notes: enter image description here

$f(z)$ can be expressed in the following way:

$f(z) = g(z) + k_1/(z-1)$

Where g(z) is analytic and $k_1$ is supposed to be the residue. However, dividing the Laurent Expansion of $e^z$ by $1/(z^2-1)$ I get:

$e^z/(z^2-1) = e(1/(z^2-1)+1/(z+1)+(z+1)/(2!(z-1))+(z+1)^2/(3!(z-1))+...$

First of all, there is a $e/(z^2-1)$ which is not analytic. Second of all, I can't seem to locate the coefficient which is the residue.

What am I missing?

Any help would be much appreciated

  • $\begingroup$ What is the name of the theorem to say that you can write $f(z)=g(z)+k_1/(z-1),$ with $g(z)$ analytic? $\endgroup$ Jun 20, 2019 at 19:21
  • $\begingroup$ I wrote it down in one of my classes of functional analysis: $\endgroup$
    – Dave A.
    Jun 20, 2019 at 19:29
  • $\begingroup$ Do you mean $...+(z-1)^2/2!+(z-1)^3/3!...$ for the expansion at $z=1$? $\endgroup$ Jun 20, 2019 at 19:30
  • $\begingroup$ I did, I see that it came out wrong. I'll fix the first post. $\endgroup$
    – Dave A.
    Jun 20, 2019 at 19:31
  • $\begingroup$ The reason I ask is because I don't buy it. In a general Laurent series, the residue is the coefficient of $(z-z_0)^{-1},$ but as J. W. Tanner is getting it, you still have the rest of the negative power terms, every one of which is not analytic at the residue point. $\endgroup$ Jun 20, 2019 at 19:32

1 Answer 1


I do not think you are calculating the Laurent series correctly. Note that any Laurent series is always taken at one point. So, while you may (and in this case, do) have multiple singularities, you can only choose one singularity at a time about which to expand your function in a Laurent series.

Your function does, indeed, have a simple pole at $z=1,$ and therefore your equation $f(z)=g(z)+k_{-1}/(z-1)$ should work. Your residue is correct. Mathematica yields $$\frac{e^z}{z^2-1}=\frac{e}{2(z-1)}+\frac{e}{4}+\frac{e(z-1)}{8}+\frac{e(z-1)^2}{48}+\dots$$ There are many ways you could obtain this. Here's how I would go about it: \begin{align*} e^z=e\cdot e^{z-1}&=e\sum_{n=0}^{\infty}\frac{(z-1)^n}{n!}, \\ \frac{1}{z+1}&=\frac{1}{z-1+2} \\ &=\frac12-\frac{z-1}{4}+\frac{(z-1)^2}{8}-\frac{(z-1)^3}{16}+\dots,\\ \frac{e^z}{z-1}&=e\sum_{n=0}^{\infty}\frac{(z-1)^{n-1}}{n!} \\ &=\frac{e}{z-1}+e+\frac{e(z-1)}{2}+\frac{e(z-1)^2}{6}+\frac{e(z-1)^3}{24}+\dots \end{align*} Finally, we multiply the two series together: \begin{align*} \frac{e^z}{(z-1)(z+1)}&=\left(\frac{e}{z-1}+e+\frac{e(z-1)}{2}+\frac{e(z-1)^2}{6}+\frac{e(z-1)^3}{24}+\dots\right) \\ &\times\left(\frac12-\frac{z-1}{4}+\frac{(z-1)^2}{8}-\frac{(z-1)^3}{16}+\dots\right) \\ &=\frac{e}{2(z-1)}-\frac{e}{z-1}\frac{z-1}{4}+\frac{e}{2}+\dots \end{align*} You see what needs to be done: collect the terms of the right power together to get the final coefficients for the various powers. You'll have more and more multiplications to do as you get further out.

  • $\begingroup$ Fantastic, that does clarify things. Thank you very much! $\endgroup$
    – Dave A.
    Jun 20, 2019 at 20:05
  • $\begingroup$ @DaveA. You're very welcome! $\endgroup$ Jun 20, 2019 at 20:09

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