Integrating a function using residues theorem Let $f(z) = \frac{1}{e^z - 1 }$. I want to compute $\int_{\gamma} f dz $ where $\gamma$ is the circle of radius $9$ centered at $0$.
Try:
I know $z = 0$ is a singularity of $f(z)$. Let $f(z) = g(z)/h(z)$ where $g(z) = 1 $ and $h(z) = e^z - 1$. We know $h(0) = 1 \neq 0 $ and $ h'(0) = e^0 = 1 \neq 0 $. Therefore, $Res( f, 0) = \frac{ g(0)}{h'(0)} = 1 $. Since $z  = 0$ is inside the curve $\gamma$, we have
$$ \int_{\gamma} f dz = 2 \pi i \times Res(f, 0) = 2 \pi i $$
is this correct?

Yes it is right. To see this explicitly,
$$\frac{1}{e^z - 1} = \frac{1}{z(1 + z/2 +z^2/3! + \dots)} = \frac{1}{z}(a_0 + a_1z + a_2z^2 + \dots)$$
By comparison, $a_0 = 1, a_1 = -1/2, \text{and}  \;a_2 = 1/12$. The analytic part of the series will of course go to $0$, leaving you to integrate $$\int_\gamma \frac{1}{z} dz = 2\pi i.$$ 
