I'm currently stuck on this problem.
Let $\gamma$ be a circle centered at zero and with a radius of 2. Find $$\int_\gamma \frac{e^\bar{z}}{z^2} \, \mathrm{d}z.$$
I saw this post which was extremely helpful for the trick converting it to an analytic system with the new radius. However, I am currently stuck as now I believe I have the integral that looks like this, and don't know where to go from here. $$\int_\gamma \frac{e^\frac{4}{z}}{z^2} \, \mathrm{d} z$$
Since$$e^{\frac4z}=1+\frac4z+\frac{4^2}{2z^2}+\frac{4^3}{3!z^3}+\cdots,$$you have$$\frac{e^{\frac4z}}{z^2}=\frac1{z^2}+\frac4{z^3}+\frac{4^2}{2z^4}+\frac{4^3}{3!z^5}+\cdots,$$and so, by the residue theorem, your integral is equal to $0$; since$$\operatorname{res}_{z=0}\frac{e^{\frac4z}}{z^2}=0.$$