I have a function that diverges like $e^{1/z}$ in the origin, and I'm integrating the function a closed contour in the complex plane. The point $z=0$ lies inside the contour, so if I integrate by residues the singularity at $z=0$ (a pole?) must be taken into account. How would you do that?


Hint. It's neither a pole nor a removable singularity. Write the Laurent series expansion of $e^{1/z}.$ This is a classic example where a function has an essential singularity (at $z=0.$)

  • $\begingroup$ If I write the Laurent expansion (as in math.stackexchange.com/questions/231133/…), it becomes a sum of infinte poles. Is that how you would compute the residue? $\endgroup$ – Jennifer Dec 29 '16 at 7:48
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    $\begingroup$ The residue would be 1. Because the contour integral around $z^{-n}$, for $n>1$ ($n$ being an integer), is zero, the residue only takes into account the $z^{-1}$ term of the Laurent series. $\endgroup$ – user361424 Dec 29 '16 at 7:50

$$e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdot\cdot\cdot, -\infty<z<\infty$$

This may also be written in the following manner.

$$e^z=\sum_{j=0}^\infty \frac{{z}^j}{j!}$$

Now, for $f(z)=e^{\frac{1}{z}}$ (which happens to have a singularity at $z=0$) we can write:

$$e^{\frac{1}{z}}=\sum_{j=0}^\infty \frac{(\frac{1}{z})^j}{j!}=\sum_{j=0}^\infty \frac{1^j}{j!\cdot z^j}$$

We know the residue would be the coefficient of the term containing the $-1^{th}$ power of $(z-z_o)$.

Coefficent of $j^{th}$ term would be:


Clearly, the term we're looking for can be obtained at $j=1$.



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    $\begingroup$ Learn to use TeX in the site and write down an answer, please. You can find a tutorial in the Meta site. $\endgroup$ – Pedro Tamaroff May 13 '17 at 9:38

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