I think that you have the correct residues at $z = -2\pm5i$. The residue at $z=3$ is wrong. It should be $$\left.\frac d{dz}\,\frac{e^{iz}}{z^2 + 4z + 29}\right|_{z=3},$$ which I get to be $\dfrac{(5i-1)e^{3i}}{250}.$Doomknightx9 said:My residue is wrong. What is the solutions and the steps to achieve it ?