Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 5 / i / Solution

Both numerator and denominator have simple zeros at zero, and $1-e^{-z}=z+O(z^2)$. Their quotient tends to one and extends holomorphically with that value. Hence the singularity is removable, and $\boxed{{\mathrm{Res}}_{0}f=0}$ because its Laurent series has no negative powers.