# Stuck on contour integral, can I use Cauchy's theorem instead?

I am trying to calculate

$$\int_{|z-2| = 3} e^{1/z}$$

I parametric the circle of radius three centered at 2 by $$\gamma(t) = 3e^{it} +2$$ and so I can instead evaluate

$$\int_\gamma f(\gamma(t))\gamma'(t) = \int_0^{2\pi}\exp\left(\frac{1}{3\exp(i\theta)+2}\right)3i\exp(i\theta)$$ and I am stuck here because I do not know how to compute the antiderivate of this. This makes me think there must be an easier way to do this

I am aware that the integrand is not defined at $$z = 0$$ is this what is causing problems? Since $$\gamma$$ is a closed path can I just say that the integral is zero, or does the fact that the integrand is not defined at $$0$$ mean that it is not analytic and hence I can not use Cauchy's theorem?

You can apply the residue theorem here. Since$$e^{\frac1z}=1+\frac1z+\frac1{2z^2}+\cdots,$$you know that $$\operatorname{res}_{z=0}\left(e^{\frac1z}\right)=1$$. Therefore, your integral is equal to $$2\pi i\times1=2\pi i$$.

• ah yes! Thank you, I had completely forgotten about the residue theorem. Dec 13, 2018 at 0:53
• Jose Carlos Santos, may I ask what is that hat which appeared on your avatar?
– Mark
Dec 13, 2018 at 0:54
• It is called “The Merlin”. Dec 13, 2018 at 1:04

I'm going to assume you aren't versed in the residue theorem yet. (Well, you implied that you were in a comment, so oh well. You can take this as an alternate answer, or this might be helpful to those in the future looking at this who aren't as well versed. As you will.)

Hint #1:

Since the integrand is not defined at $$z=0$$, and the contour encloses that singularity, you cannot use the Cauchy Integral Theorem, at least not immediately.

My recommendation would be to use the power series expansion of $$e^z$$ (plugging in $$1/z$$ for $$z$$). You can then express the integral by

$$\int_{|z-2|=3} e^{1/z}dz = \int_{|z-2|=3} \sum_{k=0}^\infty \frac{1}{k! \cdot z^k}= \sum_{k=0}^\infty \int_{|z-2|=3} \frac{1}{k! \cdot z^k}$$

You should find some pleasant surprises in that summation that simplify the process a bit.

Hint #2:

For any contour of positive orientation which is a circle of radius $$r$$ centered at the complex number $$z_\star$$, you can show

$$\int_{|z-z_\star|=r} \frac{1}{z^n} dz = \left\{\begin{matrix} 0 & \forall n \neq 1 \\ 2\pi i & n = 1 \end{matrix}\right.$$

• Thank you, I appreciate this answer very much as well because its gives me another insight into how to solve similar problems! Dec 13, 2018 at 0:57