# Solving a complex integral $\oint_L\frac{e^{1/(z-a)}}zdz$ using Cauchy's formula

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).

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

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

• Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $\frac{1}{a}$. Commented Jan 2, 2019 at 16:37
• 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 $a$. Commented Jan 2, 2019 at 16:51
• 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 ? Commented Jan 2, 2019 at 17:07

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 $$z = \infty$$.

• This is the better approach. Commented Jan 3, 2019 at 13:32