I have the integral $$\int_a\frac{1}{(z-1)(z-3)}dz$$ with the paths
- $a:[0,2\pi]\rightarrow\mathbb{C},x\mapsto2e^{it}$,
- $a:[0,2\pi]\rightarrow\mathbb{C},x\mapsto2005+73e^{it}$.
My question is:
How can I use Cauchy's Integral Formula and Theorem to find these?
I know that the formula is $f(a) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\, dz$, but how do I use it in these cases? Also isn't path $1$ the unit circle, which means that the integral is 0?
In the Cauchy integral formula, $f(z)$ is a function that is analytic on the interior of $\gamma$. The function $\frac{1}{(z-1)(z-3)}$ is analytic except at $z=1$ and $z=3$. The first path is not the unit circle, but rather the circle of radius 2 centered at the origin. The point $z=1$ is inside this circle, but $z=1$ is not. Thus, you can let $f(z)=1/(z-3)$, and $a=1$ and apply Cauchy's Integral Formula directly (and the answer will not be zero). The second path is a circle of radius $73$ centered at $2005$. It contains neither $z=1$ nor $z=3$, so $\frac{1}{(z-1)(z-3)}$ is analytic in the interior. What does Cauchy's theorem tell you about the integral?