Consider the function $$f(z) = \frac{z-2}{z^2} \sin\left(\frac{1}{1-z}\right)$$

We have that $0$ is a double pole and $1$ is a single pole (essential singularity) of $f$. It is simple to compute ${\rm Res}(f,0)$. But I'm having trouble computing ${\rm Res}(f,1)$. If I didn't mess up any calculations, so far I have: $$f(z) = (1-(z-1))\left(1+\sum_{n \geq 0}(-1)^n(n+1)(z-1)^n\right)\left(\sum_{n \geq 0}\frac{(-1)^n}{(2n+1)!(z-1)^{2n+1}}\right).$$If I only had power series, I could just use some Cauchy product to do it, but I'm stuck. Wolfram Alpha doesn't compute the residue, and don't compute the Laurent expansion either. Can someone help?

up vote 1 down vote accepted

If you consider the contour integral of f(z) for a circle with center the origin and radius R, then you see that for R to infinity, the contour integral tends to zero. This means that the sum of the residues of all the poles equals zero. Since you can easily calculate the residue at the only other pole at z =0, the answer is easily obtained from that.

  • Oh, you're right. That's quite simple, actually. Thanks! And I should've realized that the singularity is essential from my third series.. – Ivo Terek Jun 26 '15 at 16:29

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