Evaluation of the contour integral $\int_\beta \frac{e^z}{e^z-\pi} \mathrm dz$

Question

Suppose $\beta$ is a loop in the annulus $\{z:10<\left|z\right|<12\}$ that winds $N$ times about the origin in the counterclockwise direction, where $N$ is an integer. Determine the value of the contour integral $$\int_\beta \frac{e^z}{e^z-\pi} \mathrm dz$$

Well, it certainly looks like a Cauchy's theorem problem of some kind, but I'm not sure how to get it in the right format. Algebraically it doesn't seem like there's any way to turn this into the form $$\int \frac{f(z)}{(z-a)^n}\mathrm dz$$ The only idea I have is to somehow consider this in "$e^z$ space," but I'm not sure how to pull this off.

(Note: This is exam review, not homework.)

Answer 1

The function $f:z\mapsto \dfrac{e^z}{e^z-\pi}=\dfrac{e^z}{g(z)}$ is meromorphic on $\Bbb C$ and has three poles in the interior of the loop $\beta$.

They are the zeros of $g$, $z_k=\ln(\pi)+2ik\pi$, $k\in\Bbb Z$, that have a modulus lesser than 10, i.e. $z_{-1}=\ln(\pi)-2i\pi$, $z_0=\ln(\pi)$ and $z_1=\ln(\pi)+2i\pi$.

You can easily check that the other zeros of $g$ have a modulus greater than 12.

We have, for all $k\in\Bbb Z$, $\mathrm{Res}(f,z_k)=\displaystyle\lim_{z\to z_k}\ (z-z_k) f(z)=\dfrac{e^{z_k}}{g'(z_k)}=\dfrac{e^{z_k}}{e^{z_k}}=1$

By the residue theorem : $\displaystyle\int_\beta f(z) dz=2i\pi\sum_{k=-1}^1I(\beta,z_k)\mathrm{Res}(f,z_k)=6i\pi N$

Answer 2

Lets do this another way: $$\int_\beta \frac{e^z}{e^z-\pi}\,\mathrm{d}z = \int_\beta \frac{\mathrm{d}z}{1-e^{\log(\pi)-z}} = \int_\beta \frac{(z-\log(\pi))(1-e^{\log(\pi)-z})^{-1}}{z-\log(\pi)}\,\mathrm{d}z$$ I claim $f(z) = (z-\log(\pi))(1-e^{\log(\pi)-z})^{-1}$ is analytic near $\log(\pi)$, for $$f(z) = \frac{z-\log(\pi)}{1-e^{\log(\pi)-z}} = \frac{z-\log(\pi)}{-\sum_{n=1}^\infty\frac{1}{n!}(\log(\pi)-z)^n} = \frac{1}{\sum_{n=1}^\infty\frac{1}{n!}(\log(\pi)-z)^{n-1}},$$ and thus especially $f(\log(\pi))=1$.

As Girianshiido remarks, there are three relevant $\log(\pi)$'s.

Hence by Cauchy, $\displaystyle \int_\beta \frac{f(z)}{z-\log(\pi)}\,\mathrm{d}z = 3\cdot 2\pi i N$.