Ok, so I'm trying to figure out this problem. It asks to find the value of the contour integral $\dfrac{e^z}{z^2(z-\pi i)}$ on the contour $C$ shown in the following figure
I believe that in order to solve this I have to use the formula $\int f(z)=dz=\int fz(t)z'(t) dt$ There are not points in the figure, it is continuous, so I am not sure what to do about the integration limits, leave them a and b???
Break the curve $C$ up into three pieces $C_1$, $C_2$, and $C_3$, where each $C_1$ is the top loop traversed CCW, $C_2$ is the middle loop traversed CCW, and $C_3$ is the right loop traversed CCW.
Then, $\displaystyle\oint_C \dfrac{e^z}{z^2(z-\pi i)}\,dz = -\oint_{C_1} \dfrac{e^z}{z^2(z-\pi i)}\,dz + \oint_{C_2} \dfrac{e^z}{z^2(z-\pi i)}\,dz - \oint_{C_3} \dfrac{e^z}{z^2(z-\pi i)}\,dz$.
The function $\dfrac{e^z}{z^2(z-\pi i)}$ has a double pole at $z = 0$ and a single pole at $z = \pi i$. So, the interior of $C_1$ has only one pole at $z = \pi i$, the interior of $C_2$ has a double pole at $z = 0$, and the interior of $C_3$ has no poles.
Now, apply the residue theorem to calculate the integrals over $C_1$, $C_2$, and $C_3$.
To calculate the integrals, first use partial fractions to get $\dfrac{1}{z^2(z-\pi i)} = \dfrac{1/\pi^2}{z} + \dfrac{i/\pi}{z^2} - \dfrac{1/\pi^2}{z-\pi i}$.
Therefore, $\displaystyle\oint_{C_i}\dfrac{e^z}{z^2(z-\pi i)}\,dz = \dfrac{1}{\pi^2}\oint_{C_i}\dfrac{e^z}{z}\,dz + \dfrac{i}{\pi}\oint_{C_i}\dfrac{e^z}{z^2}\,dz - \dfrac{1}{\pi^2}\oint_{C_i}\dfrac{e^z}{z-\pi i}\,dz$.
Applying Cauchy's Integral formula, you have that $\displaystyle\oint_{C_i}\dfrac{f(z)}{(z-a)^{n+1}}\,dz = \dfrac{2\pi i f(a)}{n!}$ if $a$ lies in the interior of the curve $C_i$, and $0$ otherwise. Can you finish the problem from here?