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!