I recently came across the exercise of integrating
$\int_{\Gamma:|z|=1}\frac{e^z-1-z}{z^2}dz,$
where, naturally, $z\in\mathbb{C}$.
The first thing I thought of was using Cauchy's integral formula
$f(z_0)=\frac{1}{2\pi i}\int_\Gamma\frac{f(z)}{z-z_0}dz,$
along with its derivative
$f^{(n)}(z)=\frac{n!}{2\pi i}\int_\Gamma\frac{f(\xi)}{(\xi-z)^{n+1}}d\xi.$
The way I proceeded was by letting $f(\xi)=e^\xi-1-\xi$, implying that $f'(\xi)=e^\xi-1$, and $f'(0)=0$. Hence,
$\begin{align} 2\pi i\cdot f'(z)&=\int_\Gamma\frac{f(\xi)}{(\xi-z)^2}d\xi\\ &\Longrightarrow\int_\Gamma\frac{e^\xi-1-\xi}{\xi^2}d\xi=0, \end{align}$
whenever $z=0$.
Is this a correct line of thought? The only thing that I cannot wrap my head around is the fact that it seems as though the value of the integral does not necessarily depend on the contour $\Gamma$. Thanks in advance!