How to evaluate an integral using residue theorem, when having an imaginary pole at the denominator?

Question

I was trying to solve this integral using Cauchy residue theorem:

$\oint_{z=|3|}\dfrac{e^{iz}}{z^2(z-2)(z+5i)}dz$

I know that there are three poles, a pole of order 2 at $z=0$, a simple pole at $z=2$ and a pole at $z=-5i$, I have not encountered an integral like this before, where there is an imaginary pole, and I started recently studying complex analysis, so excuse my ignorance. My question is, do I proceed with the problem calculating the residues at $0$ and $2$, ignoring the $-5i?$ or is there something I am missing?

Answer 1

There's nothing special about the fact that $ -5i $ isn't real. However, it's not inside the circle you're integrating on, so its winding number is 0. Thus, by the residue theorem, it makes no contribution to the integral and can therefore be ignored, but not because it's non-real.

Answer 2

While it is true that you can calculate the residue at $-5i$ in the same way as you wound at a real point, the important thing here is that $\lvert -5i \rvert = 5>3$, so the point lies outside the contour anyway and should be ignored.