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I need to calculate the following contour integral using residue theory.

$z \in \mathbb{C}$

$f(z)=\exp(-1/z) \sin(1/z)$

$\oint_C f(z) dz$

$C: \left | z \right |=1$

The difficult points I detected include only z=0.

But i'm a bit stuck at calculating the residue of f at z=0. (I know the answer is -1)

I know that the residue is the coefficient of the 1/z term in the laurent series. However, I don't know how to create the laurent series of this function quickly,

I've been playing around with a substitution: $z=1/\xi$ but that didn't help me so far.

Its probably really easy but I'm new to the subject. So any tips are appreciated and the one that helps me to find the solution myself will be accepted as an answer.

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1  
Euler's formula, $\sin w = \frac{1}{2i}(e^{iw} - e^{-iw})$. –  Daniel Fischer Oct 24 '13 at 20:00
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If you know the answer, using the Residue Theorem you know what the residue should come out to be. To get the answer, write out the Laurent series for exp(-1/z) (easy enough) and the one for sin(1/z). Then just multiply them together. You needn't find all the terms, just the ones whose product gives the proper exponent. –  mathematics2x2life Oct 24 '13 at 20:02

2 Answers 2

up vote 3 down vote accepted

$$e^{-1/z} \sin \frac1z = \left(1 - \frac1z + \ldots\right)\left(\frac1z+\ldots\right).$$

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Isn't that specific expansion valid only around (1/z) = 0. here 1/z goes to infinity (because it is around z=0) ? –  tgoossens Oct 24 '13 at 20:11
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@tgoossens: that is just plugging $1/z$ into the power series. It is good everywhere but $z=0$ –  robjohn Oct 24 '13 at 20:12
    
Ok. I looked at it for a while. I guess this is valid because the radius of convergence of the exponential function is infinite. And the sine can be written as a sum of exponential functions so the same argument is valid. Is this a correct justification of your answer? –  tgoossens Oct 24 '13 at 20:25
    
Yes, $e^z$ and $\sin z$ are entire functions, i.e. holomorphic on $\mathbb{C}$. –  njguliyev Oct 24 '13 at 20:28

Hint: Note that using the substitution $z=\frac1w$, we get $$ \oint_C\exp(-1/z)\sin(1/z)\,\mathrm{d}z=\oint_C\exp(-w)\sin(w)\frac{\mathrm{d}w}{w^2} $$ The $-1$ from the $\mathrm{d}z=-\frac{\mathrm{d}w}{w^2}$ is cancelled by the change in direction of the contour.

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