Consider $$\int_C\frac {{e^z}^t}{(z^2+1)^2} dz,$$ where $t>0$ and $C$ is the circle $|z| = 3$.
I am able to solve it when $z=i$ and $z=-i$ are two simple poles, but cannot proceed when the same poles with order $2$. Please explain how to proceed in such cases.
You can apply partial fraction decomposition:$$\frac1{(z^2+1)^2}=\frac14\left(\frac i{z+i}-\frac1{(z+i)^2}-\frac i{z-i}-\frac1{(z-i)^2}\right)$$and therefore$$\frac{e^{zt}}{(z^2+1)^2}=\frac14\left(\frac{ie^{zt}}{z+i}-\frac{e^{zt}}{(z+i)^2}-\frac{ie^{zt}}{z-i}-\frac{e^{zt}}{(z-i)^2}\right).$$And now you can apply Cauchy's integral formula four times to compute your integral.