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I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:

$$\oint_{L}{\frac{e^{\frac{1}{z-a}}}{z}}dz$$ where $$L=\{z\in\mathbb{C}:|z|=r\}$$ for some $r\neq|a|$.

So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula: $$\oint_{L}{\frac{e^{\frac{1}{z-a}}}{(z-0)}}dz = \frac{2\pi i}{e^{1/a}}$$ If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).

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2 Answers 2

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Hint

I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.

For a radius $r > \vert a \vert$, you get

$$\oint_{L}{\frac{e^{\frac{1}{z-a}}}{z}}dz =2i\pi \left(\frac{1}{e^{1/a}}+\mbox{Res}(f,a)\right)$$

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  • $\begingroup$ Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $\frac{1}{a}$. $\endgroup$
    – math101
    Commented Jan 2, 2019 at 16:37
  • $\begingroup$ I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $\sum_{-\infty}^\infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$. $\endgroup$ Commented Jan 2, 2019 at 16:51
  • $\begingroup$ I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ? $\endgroup$
    – math101
    Commented Jan 2, 2019 at 17:07
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See this answer. Instead of taking the residue at $z = a$, take the residue at $z = \infty$, which is the value of $-e^{1/(z - a)}$ at $z = \infty$.

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  • $\begingroup$ This is the better approach. $\endgroup$
    – Szeto
    Commented Jan 3, 2019 at 13:32

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