$\int \frac{(e^z)}{z-\pi i} dz$, if C is the ellipse |z - 2| + |z+2| = 6

Question

$$\oint_{C} \frac{e^z}{z-\pi i}\ \mathrm dz$$ If $C$ is the ellipse $|z - 2| + |z+2| = 6$, how is it $0$? I can find the singular point in $2$ and $-2$.

Answer 1

Graph of contour and pole at $z=i\pi$

Perhaps this visual representation can help you intuitively understand why

$$\oint_{C} \frac{e^z}{z-\pi i}\ \mathrm dz=0$$ Where $C$ is the ellipse $|z - 2| + |z+2| = 6$

Answer 2

First, consider the set $S=\{z\in\mathbb{C}:|z-2|+|z+2|<2\sqrt{13}\}$. This set can be shown to be simply connected.

Let $z\in C$

$\Rightarrow z\in\mathbb{C}$

Suppose $z-\pi i=0$

$\Rightarrow z=\pi i\Rightarrow |z-2|+|z+2|=|\pi i-2|+|\pi i+2|=2\sqrt{\pi^{2}+4}>2\sqrt{9+4}=2\sqrt{13}\Rightarrow z\notin C$

Thus by the contrapositive, $z\in C\Rightarrow z-\pi i\neq 0$

$\Rightarrow z-\pi i\neq0$

Thus, $\frac{e^z}{z-\pi i}$ is holomorphic on $S$.

Let $z$ be in the curve.

$\Rightarrow |z-2|+|z+2|=6=2\sqrt{9}<2\sqrt{13}$

$\Rightarrow z\in S$

Thus, the curve is inside $S$, and as $\frac{e^z}{z-\pi i}$ is holomorphic on $S$, so $\oint_C {\frac{e^z}{z-\pi i}}\, {dz}=0$