Cauchy Integral Formula How can I apply the Cauchy Integral Formula if given a contour, two singular points are inside it? for example, how could I evaluate $\int \frac{dz}{z(z-2)}$ given $C: z = 3e^{i\theta}$ 
 A: By the Cauchy-Goursat theorem, deforming a contour continuously within a region where the integrand is analytic won't change the value of the integral. Your contour deforms to two circles, each enclosing just one of the poles (discontinuities) of the integrand, connected by a line segment. That line segment is traversed backwards once and forwards once, so cancels itself in the integral. Therefore you can just add the integrals around little circles around each pole. (This argument works in generality when you are integrating a function that's holomorphic away from a finite set of points and is not special to your particular integral.)
A: I just want to add something to the solutions given here. Of course it is possible to use the residue theorem for the solution, but you can also do it directly by using CIT.
At first you have to do a partial fractioning with $f(z)=\frac{A}{z}+\frac{B}{z-2}$. $A$ and $B$ are the residues of $f$ in $z=0$ and $z=2$ (you can compare the coefficients as well). You get $A=-\frac{1}{2}$ and $B=\frac{1}{2}$. The partial fractioning of $f$ ist now $f(z)=\frac{-\frac{1}{2}}{z}+\frac{\frac{1}{2}}{z-2}$.
Using CIT, you have $f(a)=\frac{1}{2\pi\mathrm{i}}\oint_C \frac{f(z)}{z-a}\mathrm{d}z$. The contour $C$ consists every pole so that you can sum up the values of the separate integrals returned by the CIT. For using the CIT you just pick the numerator function, for the first fraction $f(z)=-\frac{1}{2}$. It is now clear that $f(a)=f(0)=-\frac{1}{2}$. Multiplying $f(0)$ by $2\pi\mathrm{i}$, you get the value of the contour integral.
The result is zero, because of $2\pi\mathrm{i}(-\frac{1}{2}+\frac{1}{2})=0$. This is of course the same value that you can get by applying the residue theorem. 
A: Take little circles around each of the poles, apply CIT to each of these and sum...
Note that when you join the outer countour $\,C:=\{z\in\Bbb C\;;\;|z|=3\}\,$ with these little contours around the poles, the integrals on the joining lines vanish (once in one direction in, the other direction out!) , so at the end you get your integral on your wanted contour.
The solution here is
$$2\pi i\left(Res_{z=0}(f)+Res_{z=2}(f)\right)\;\;,\;\;\text{with}\;\;\;f(z)=\frac{1}{z(z-2)}$$
