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In our lecture we defined the residue of a function $f(z)= \sum_{-\infty}^{\infty}a_nz^n$ as the coefficient $a_{-1}$ of the Laurent series.

We wanted to calculate the residue of $z\exp(\frac{1}{1-z})$.
In class they did it by substituting $z = 1+h$. So we get $(1+h)\exp(\frac{-1}{h}) = (1+h)(1-\frac{1}{h} + \frac{1}{2h^2} -...)$ The coefficient $a_{-1} = -1 + \frac{1}{2} = - \frac{1}{2}$

If I do the exact same calculation, but substitute $z = 1-h$, I get $a_{-1} = + \frac{1}{2}$.

Which of those solutions is correct? Is any of those solutions correct? I am a little bit confused right now.

Edit: I am dealing with the residude at $z=1$.

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Both approaches are correct. If you do the substitution $z=1-h$, this minus induces a sign change in those $a_n$'s for which $n$ is odd and, in particular, for $n=-1$. Therefore, using this second substitution, the conclusion is, again, that the residue is $-\frac12$.

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  • $\begingroup$ Could you explain what the OP is doing? To me it seems like the residue should be $0$. We have that $1/(1-z) = \sum_{n\geq0} z^n$ and $e^z = \sum_{n\geq0} z^n/n!$ so I don't see how the composition would produce a nonzero coefficient for $1/z$. $\endgroup$ – Tai Jul 28 '17 at 18:31
  • $\begingroup$ @TaisukeYasuda Although the OP didn't say so, I assumed that we were dealing with the residue at $1$ here. $\endgroup$ – José Carlos Santos Jul 28 '17 at 18:34
  • $\begingroup$ Got it, thank you! $\endgroup$ – Tai Jul 28 '17 at 18:35
  • $\begingroup$ I was dealing with the residue at $z = 1$. $\endgroup$ – Limechime Jul 28 '17 at 18:35

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