I want to understand why I can't use Cauchy Integral Formula for the following problem:
$$\int_C \frac2{z^2 -1}dz\text{ on the contour } |z-1|=\frac12$$
Now it says that I need $f$ to be analytic everywhere inside and on a simple closed contour. My contour is surely simple and closed. To check that $f$ is analytic, I need to use the Cauchy-Riemann equations I imagine – but let's ignore that since it is known to be analytic(two polynomials are analytic?)
So this has one singularity at $z=1$, so then I want to use:
$$\int_C\frac{2}{z^2-1}=\int_C \frac{2}{(z^2-1)(z-1)} dz=2\pi i f(1)$$
But $f(1)$ makes this diverge, what's the deal?
Best Answer
You should specify what your $f$ is. Remember that our goal is to rewrite the integral in the form $\displaystyle \int_C \frac{f(z)}{z-a} \ dz$.
Since $z^2-1 = (z+1)(z-1)$, then $\displaystyle f(z) = \frac{2}{z+1}$ is a good choice. Notice that $f$ is analytic everywhere on and inside your contour, as required (in particular, $z=-1$ is the only point at which $f$ is not analytic). So your integral is:
$$\int_C \frac{2}{z^2-1} \ dz = \int_C \frac{ \frac{2}{z+1}}{z-1} \ dz = \int_C \frac{f(z)}{z-1} \ dz$$
And at this point you can apply Cauchy's integral formula to finish up.