Complex Integral Cauchy Integral Formula Im trying to calculate 
$$\int_{\gamma}\frac{e^{iz}}{z^2} dz$$  where $\gamma(t)=e^{it}, 0\le t \le 2\pi$
My problem is that $z^2$ in the denominator.
If the function was  $\frac{e^{iz}}{z}$ as $f(z)=e^{iz}$ is holomorphic 
$\int{_{\gamma}\frac{e^{iz}}{z}dz}=2\pi if(0)=2\pi i$
Obs: I dont know Residue Theorem 
Thanks!
 A: Doing a Taylor expansion of $e^{iz},$ resulting in a Laurent expansion of $\frac{e^{iz}}{z^2},$ we have
$$
\oint_\gamma \frac{e^{iz}}{z^2} dz 
= \oint_\gamma \frac{1+iz+O(z^2)}{z^2} dz
= \oint_\gamma \left( \frac{1}{z^2} + i\frac{1}{z} + O(z) \right) dz
$$
Here $O(z)$ is holomorphic.
It's now easy to verify now that
$$\oint_\gamma \frac{1}{z^2} dz = 0, \quad \oint_\gamma \frac{1}{z} dz = i\ 2\pi \quad \oint_\gamma O(z) dz = 0$$
Thus,
$$\oint_\gamma \frac{e^{iz}}{z^2} dz = i \cdot i\ 2\pi = -2\pi$$
A: Note that from the Cauchy Integral Formula
$$f(z)=\frac1{2\pi i}\oint_{|z'-z|=1}\frac{f(z')}{z'-z}\,dz'$$
we can express the first derivative of $f(z)$, $f'(z)$, as 
$$f'(z)=\frac{1}{2\pi i}\oint_{|z'-z|=1}\frac{f(z')}{(z'-z)^2}\,dz'\tag 1$$
Setting $z=0$ in $(1)$ yields
$$f'(0)=\frac{1}{2\pi i}\oint_{|z'|=1}\frac{f(z')}{z'^2}\,dz'\tag2$$
Now, letting $f(z)=e^{iz}$ with $f'(0)=i$ in $(2)$ reveals
$$i=\frac{1}{2\pi i}\oint_{|z'|=1}\frac{e^{iz'}}{z'^2}\,dz'$$
whereupon solving for the integral of interest we find
$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z'|=1}\frac{e^{iz'}}{z'^2}\,dz'=-2\pi}$$
