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I am trying to calculate the residue of $ze^{\frac{1}{z}}$, here's what I got:

We have a singularity at $z=0$.

We know that $e^w=\sum_{n=0}^\infty \frac{w^n}{n!}$ so $e^{\frac{1}{z}}=\sum_{n=0}^\infty \frac{(\frac{1}{z})^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!z^n}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\frac{1}{4!z^4}+...$

so

$ze^{\frac{1}{z}}= z(1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\frac{1}{4!z^4}+...)=z+1+\frac{1}{2!z}+\frac{1}{3!z^2}+\frac{1}{4!z^3}+...$.

So $z=0$ is an essential singularity as we have no negative terms of $(z-z_0)=(z-0)=z$.

Now, I thought that the residue was given by the coefficient of $\frac{1}{z}$ which in this case would be $\frac{1}{2!}$. However, when I checked the answer was 1 and I have no idea why. Is my idea of what a residue in this case is flawed? I have watched a lot of videos and read up online but it seems to all support what I thought.

Thank you in advance!

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    $\begingroup$ How did you check your answer? $\endgroup$
    – Severin Schraven
    Commented May 1 at 7:32
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    $\begingroup$ Your approach looks correct to me. $\endgroup$
    – Severin Schraven
    Commented May 1 at 7:34
  • $\begingroup$ Thank you for your comments! This question was a practise question provided to me by my university, and I checked my answer against theirs, where they said the Res = 1. $\endgroup$
    – Luke
    Commented May 1 at 7:42
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    $\begingroup$ If the question asks about the residue at $z=0$, then this seems to be a mistake in the solutions. $\endgroup$
    – Severin Schraven
    Commented May 1 at 7:45
  • $\begingroup$ Yes, the question asks for it at z=0, thank you I suppose there was a mistake. $\endgroup$
    – Luke
    Commented May 1 at 8:20

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