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I have to find poles and residues as functions of a for the following function:

$\frac{1}{e^z-a}$

But I'm not sure where to start. I know how to find poles and residues but not when I'm given an arbitrary function with an unknown, a, in this case. Any ideas?

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  • $\begingroup$ Do you know what the poles and residues are for $a\in\{0,1\}$? $\endgroup$
    – K B Dave
    Mar 21, 2018 at 2:28
  • $\begingroup$ The poles would be ln(a)? What about residues? $\endgroup$
    – mathwiz97
    Mar 21, 2018 at 2:30
  • $\begingroup$ It's not clear to me whether you interpreted my hint correctly, so here's different one: the DLMF lists the partial fraction decomposition of the function $\mathrm{csch}\, z$. $\endgroup$
    – K B Dave
    Mar 21, 2018 at 3:57
  • $\begingroup$ @KBDave in what way does that relate? I'm struggling to see that! $\endgroup$
    – mathwiz97
    Mar 21, 2018 at 4:12
  • $\begingroup$ Sorry, I'll try and write a more lucid answer $\endgroup$
    – K B Dave
    Mar 21, 2018 at 4:43

2 Answers 2

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It is clear that there are countable many poles , each of order 1 for $a\neq 0$. [ Some very basic knowledge of logarithms gives this. To see that poles are of order 1 just check that the derivative of $e^{z} -a$ does not vanish at $z_0$ if $e^{z_0} =a$]. Now the residue of $\frac 1 {e^{z}-a}$ at any pole $z_0$ is given by the limit as $z\to z_0$ of $(z-z_0)\frac 1 {e^{z}-a}$ which, by L'Hopital's Rule is $\frac 1 {e^{z_0}} =\frac 1 a$. [I leave it to you to write down the poles explicitly using logarithms].

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Suppose $a\neq 0$. Let $f(z)=\tfrac{1}{\mathrm{e}^z-1}$. Then $$\frac{1}{\mathrm{e}^z-a}=a^{-1}f(z-\ln a)$$ so it suffices to consider the residues of $f$ alone. The partial fraction decompositions in the DLMF tell us what to expect:

$$f(z)=-\frac{1}{2}+\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2+(2n\pi)^2}\text{.}$$ That is, at each zero of $\mathrm{e}^z-1$ there is a simple pole of residue $1$. To show this, it suffices to think about the behavior near $z=0$ since $f(z)$ is periodic. Note that the Maclaurin expansion of $\mathrm{e}^z-1$ starts off as $$\mathrm{e}^z-1=z+O(z)\text{.}$$ Taking the reciprocal of this expression, the Laurent expansion of $f(z)$ must start off as $$f(z)=\frac{1}{z}+O(1)\text{.}$$

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