I want to evaluate the integral $$\int_{\left | z \right |=3} \frac{e^{z}-1}{z(z-1)(z-i)}\,dz$$ by using the following theorem:
If a function $f$ is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed curve $C$, then $$\int_{C}f(z) dz=2\pi i\text{Res}\left ( \frac{1}{z^{2}}f(\frac{1}{z}), 0 \right ).$$
Since $$\frac{1}{z^{2}}f(\frac{1}{z})=\frac{1+\frac{1}{2!z}+\frac{1}{3!z^{2}}\cdot \cdot \cdot }{1-(1+i)z+iz^{2}},$$ so I did long division, but the coefficient of 1/z(the residue of $\frac{1}{z^{2}}f(\frac{1}{z})$) is $0$. I know the method of calculating the residues at each singularities $0$, $1$, $i$, respectively. This method gives me the answer $2\pi i(0+\frac{e-1}{1-i}+\frac{e^{i}-1}{i(i-1)})$ which is not $0$.
Why answers are different?? Why does not hold long division method? Any help please.