> 
> 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 
>  $$\int\_{C}f(z) dz=2\pi i\text{Res}\left ( \frac{1}{z^{2}}f(\frac{1}{z}), 0 \right ).$$
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I want to calculate this integral by calculating the residue at 0 of $\frac{1}{z^{2}}f(\frac{1}{z})=\frac{1+\frac{1}{2!z}+\frac{1}{3!z^{2}}\cdot \cdot \cdot }{1-(1+i)z+iz^{2}}$
So I did long division but the coefficient of 1/z 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.