I am having a little bit of trouble with the following:
$$\int_{\gamma}\frac{z^2-1}{z^2+1}dz$$ where $\gamma$ is a circle of radius $2$ centered at 0. I am trying to separate this or simplify it into the form in which we can maybe apply Cauchy's differentiation formula but this isn't working. I did get this:
$$\frac{z^2-1}{z^2+1} = 1-\frac{2}{z^2+1}$$ but this doesn't seem to take me anywhere. Any hints would be helpful.
Edit:
Would it be possible to do this:
$$\frac{z^2-1}{z^2+1} = 1-\frac{2}{z^2+1}= 1-\frac{\frac{2}{z+i}}{z-i}$$ and then just apply cauchy's integral formula with $z_0=i$? and $f(z)=\frac{2}{z+i}$?
Using this, we would have
$$ \frac{2}{i+i}2\pi i + \frac{2}{-i-i}2\pi i = 0 $$ as the answer?
$$ \int_\gamma 1\,dz=0 $$ because $\gamma$ returns to its starting point.
Next we have $$ z^2+1 = (z-i)(z+i). $$
So the integral should involve the sum of residues at $\pm i$, since $\gamma$ winds once around each of those two points.
PS: Your proposal to apply Cauchy's formula at $i$ to the function $$ 1-\frac{\frac{2}{z+i}}{z-i} $$ would work if not for the fact that the numerator $\dfrac2{z+i}$ also has a pole inside the curve $\gamma$. You need to take the residue at that point into account as well.