Using Cauchy Integral Formula $\int_C \frac2{z^2 -1}dz$

Question

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?

Answer 1

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.

Answer 2

Use Partial fractions to solve $\frac{1}{z^2 -1} = \frac{A}{z-1} +\frac{B}{z+1}$..Find the value of A and B and substitute and multiply with 2 since our numerator is 2.And apply Cauchy integral formula for both terms..the second will get 0 since -1 will not be in the disc |z-1|=1/2..Hence the solution..