I am trying to use local Cauchy integral formula to find the value of
$\int_{ \vert z \vert=2}(z+1)^{-2}\ e^z\ dz.$
So what I am having some troubles when trying to make this integral in the form
$\int_{\gamma} \frac{f( \zeta)}{\zeta-z} dz$, so could any one please help me with that.
Thank you in advance.
If you reduce the integral in this form $\int_{\gamma} \frac{f( z)}{z-\alpha} dz$ then $f( z)=\frac{e^z}{z+1}$ is not analytic in $|z|=2$. You can use the cauchy's integral formula in derivative form $$f^n(z)=\frac{n!}{2\pi i}\oint_C \frac {f(w)dw}{(w-z)^{n+1}} \,ds,$$ where $C$ is simple closed curve in which $f(w)$ in analytic and $z$ is interior to $C$.
Therefore $$\int_{|z|=2} \frac {e^z}{(z+1)^2}=\frac{2\pi i}{1!}e^{-1}=\frac{2\pi i}{e}.$$ Here note that $f(w)=e^w$.