I understand there are many questions regarding this topic (Complex integral where poles are outside of the given region)
I just wanted to double check with everyone what I gathered so far:
$$f(z_0) = \frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z-z_0}dz = 0, \\ $$
if $z_0$ lies outside of the given contour and there exists a closed loop around the pole where the loop is analytic everywhere (this is my understanding of holomorphic).
I think this is true due to the path independence but I can't prove it exactly. Can anyone please help?
One specific example will be $$\int_C \frac{ze^z}{2z-3}dz$$ where C be the circle $|z-1.5| = 2$ traversed once in the positive sense.
I calculated that $z_0 = 3/2$ does not lie inside of the curve ($z$ intersects with the real axis at $\pm \sqrt{1.75}$).
Hence, I claim according to the above assumption because the pole does not lie inside of the region of interest ($C$), the integral is $0$.
I stated w/o residue thrm because we didn't learn it. I see many answers use residue thrm in their answer so it will be great if you could explain it w/o residue thrm!
The line integral of a holomorphic function $f$ over the boundary of a domain where $f$ is holomorphic equals zero. Your function $\frac{f(z)}{z-z_0}$ is holomorphic inside the domain, so the integral is zero.