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Question: How to find the residue of an even function at a pole of order $~n~?$

I know how to find the residue of a function depending on the nature of the singularity. But the above question stuck me. I am unable to figure out the solution of this particular question.

What I done till now: Mathematically if I describe this question I got,

Let $~f(z)~$ be an even function and has a pole of order $~n~$ at $~z=z_0~$. Then what is the value of $~\text{Res}\left(f,z_0\right)~?$

Now let $~f(z)=\frac{g(z)}{(z-z_0)^n}~,$ where $~g(z)~$ be an analytic funtion, then

$$~\text{Res}\left(f,z_0\right)~=\frac 1{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left[(z-z_0)^nf(z)\right]=\frac 1{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left[g(z)\right]~.$$

Then what ?

Where I have to use the property of an even function ?

Any one please help.

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  • $\begingroup$ What do you mean by "even" in this context? If it's the usual "$f(-z) = f(z)$", then do you also mean to take $z_0=0$? $\endgroup$ – MPW Aug 21 at 12:19
  • $\begingroup$ $(z)=f(-z)\Rightarrow a_{2n-1}=0$ for any $n\in\mathbb{Z}$ where $f(z)=\sum_{n=-\infty}^{\infty} a_{n}z^{n}$ $\endgroup$ – user723846 Aug 21 at 12:28
  • $\begingroup$ @MPW: even function means $~f(z)=f (-z)~.$ But I am unable to understand why we take $~z_0=0~?$ $\endgroup$ – nmasanta Aug 21 at 13:01
  • $\begingroup$ @Gio: I am unable to get your indication. Can you elaborate ? $\endgroup$ – nmasanta Aug 21 at 13:05
  • $\begingroup$ If $f(z)=\sum_{n=-\infty}^{\infty} a_{n}(z-c)^{n}$ for some $(a_{n})_{n\in\mathbb{Z}}$ then for any $(b_{n})_{n\in\mathbb{Z}}$ such that $f(z)=\sum_{n=-\infty}^{\infty} b_{n}(z-c)^{n}$ one can infer that $a_{n}=b_{n}$ for any $n\in \mathbb{Z}$, by Laurent Theorem. Let $b_{n}=(-1)^{n}a_{n}$. $\endgroup$ – user723846 Aug 21 at 14:16
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If $z_0 = 0$ then the residue must be zero: In the Laurent series $$ f(z) = \sum_{k=-n}^\infty a_k z^k $$ of $f$ at $z=0$ all odd-indexed coefficients are zero because $f$ is even. It follows that $$ \operatorname{Res}(f, 0) = a_{-1} = 0 \, . $$

If $z_0 \ne 0$ then nothing can be said about the residue: For arbitrary complex numbers $c_1, \ldots, c_n$ with $c_n \ne 0$ the function $$ f(z) = \sum_{k=1}^n \left( \frac{c_k}{(z-z_0)^k} + \frac{c_k}{(-z-z_0)^k} \right) $$ is even and has a pole of order $n$ at $z_0$ with $\operatorname{Res}(f, z_0) = c_1$.

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