cauchy's integral formula with singularities I'm trying to evaluate
$$\int_{\gamma} \frac{e^z}{z^m(1-z)}dz$$ 
where $\gamma$ is the boundary of $D(\frac{1}{2},1)$.
I can't apply Cauchy's integral formula, since the function has its singularities inside the circle. How can I proceed?
 A: You do not need the residue theorem. You can decompose your integrand into
$$ \int_{\gamma} e^{z} \left( \frac{1}{z} + \frac{1}{z^2} + \ldots + \frac{1}{z^m} + \frac{1}{1 - z} \right) \, dz.$$
To justify the partial fraction decomposition, notice that by adding and subtracting terms,
\begin{align}
\frac{1}{z^m (1 - z)} & = \frac{(z^{m-1} + \ldots + z + 1)(1 - z) + z^m}{z^m(1 - z)} = \frac{z^{m-1}(1 - z) + \ldots + (1 - z) + z^m}{z^m(1 - z)} \\
& = \frac{1}{z} + \ldots + \frac{1}{z^m} + \frac{1}{1 -z} \end{align}
Now, you can calculate each piece of the integral using the generalized Cauchy integral formula, namely that
$$ f^{(k)}(z_0) = \frac{k!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}} \, dz$$
for $f(z)$ analytic. I'll leave it to you to take it from here.
EDIT: Justified the partial fraction decomposition.
A: You can use the Residue theorem.
For every $a \in D(\frac{1}{2},1)$ we have $$\int_\gamma f(z) ~dz= 2\pi i\sum \text{Res}(f, a_k),$$
Where the $a_k$ are the residues of the poles of $f$ inside $\gamma$. The residues of $f$ are given by the coefficients of $z^{-1}$ in the Laurent Series expansion of $f(z)$ at each of the singularities.
In this case we have two singularities, a simple pole at $z=1$ and a pole of order 5 at $z=0$.
There are in fact closed form identities for determining residues of $f$ at these sorts of poles.
For a simple pole we have,
$$\text{Res}(f, a) = \lim_{z \to a}(z-a)f(z),$$
and for the pole of order n we have,
$$\text{Res}(f, a) = \frac{1}{(n-1)!}\lim_{z \to a}\frac{d^{n-1}}{dz^{n-1}}((z-a)^nf(z))$$
You should now be able to be able to compute the integral, by substituting in the derived values for the residues.
