I want to solve this integral integral
$$\int_{\gamma} \frac{e^{iz}}{z^2},\quad \gamma(t)=e^{it}\text{ and } 0 \leq t\leq 2\pi$$
My attempt:
$$f(\gamma(t))=\frac{e^{ie^{it}}}{{(e^{it})}^2}={(e^i)^{e^{it}-2t}} \text{ and } \gamma'(t)=ie^{it}$$
So $$\int_{\gamma}\frac{e^{iz}}{z^2}= \int_{o}^{2\pi}{(e^i)^{e^{it}-2t}} \cdot ie^{it}=i\int_{o}^{2\pi}{(e^i)^{e^{it}-t}}$$
from here I don't know how to continue, it's probably something basic, but I can't find it.
This is best solved by the residue theorem, you have an enclosed pole in the contour. Develop the exponential in its Taylor expansion and observe the residue value at $z = 0$.
$$e^{iz} = 1 + (iz) + \frac{(iz)^2}{2!} + ..$$ Hence $$ \frac{e^{iz}}{z^2} = \frac{1}{z^2}+ \frac{i}{z} - \frac{1}{2!} + ..$$ Hence, the residue at zero is just $i$; there is a simple pole at $z= 0$.
The value of the integral is $2\pi i$ times the sum of the enclosed residues so it's $-2\pi$.
