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

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

Let $zin C$

$Rightarrow zinmathbb{C}$

Suppose $z-pi i=0$

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

Thus by the contrapositive, $zin CRightarrow z-pi ineq 0$

$Rightarrow z-pi ineq0$

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

Let $z$ be in the curve.

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

$Rightarrow zin S$

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

Graph of contour and pole at $z=ipi$

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 $|z - 2| + |z+2| = 6$