How to calculate $\int_{C_a} \frac{e^z}{(z-a)(z-b)}dz$. Here, $C_a$ is a "small" circle about $a$ so that $b$ is outside of $C_a$.
Actually, if I can prove the following statement, I can calculate $\int_{C_a} \frac{e^z}{(z-a)(z-b)}dz$ very easily.
On the third edition of Ahlfors' Complex Analysis, page 151 it states: For simple poles the method is even more immediate, for then the residue equals the value of the function $(z-a)f(z)$ for $z=a$.
I want to make the statement more clear. Let $g(z)=(z-a)f(z)$. I should prove $\frac{1}{2\pi i}\int_{C_a}f(z)dz=g(a)$. But I don't know how to prove it.
Cauchy's integral formula can be applied to $h(z) = e^z/(z-b)$, which is holomorphic in disk containing $C_a$: $$ \int_{C_a} \frac{e^z}{(z-a)(z-b)} \, dz = \int_{C_a} \frac{h(z)}{z-a} \, dz = 2 \pi i h(a) = \frac{2 \pi i e^a}{a-b} \, . $$