contour integrals parametrising and solving Use Cauchy's integral formula to compute the following:
$$\int \limits_{\Gamma} \frac{e^{-z}}{z-1}dz$$ where $\Gamma$ is the square with parallel sides to the axes, centre $i$ and side length $5$ traversed in the anticlockwise direction.
So do I split this integral into four parts? Would this parametrisation work:
Parametrise $\Gamma$: Define $\gamma (t)=p+t(q-p)$ where $p,q \in \mathbb{C}$ and $t \in [-1,0]$?

No, do not split up the integral. Since $f(z) = e^{-z}$ is analytic inside and on $\Gamma$, and since $z_0 = 1$ lies inside $\Gamma$, your integral, which is $\int_{\Gamma} \frac{f(z)}{z - z_0}$, is equal to $2\pi i f(z_0) = \frac{2\pi i}{e}$ by Cauchy's integral formula.
