$\int\limits_{\gamma} \frac{z}{(z1)(z2)}$, $\gamma(\theta) = re^{i\theta}$
$\int\limits_{\gamma} \frac{z}{(z1)(z2)}$, $\gamma(\theta) = re^{i\theta}$, $2 < r < \infty$
For $0 < r < 2$, we can use Cauchy's integral formula and choose our holomorphic function to be $f(z) = \frac{z}{z  2}$ since $z = 1$ is the only pole, but if $r > 2$, then both poles $z = 1$ and $z = 2$ are inside the contour so we can use partial fractions to get our integral into a form to use Cauchy's integral formula, but the solution says that because of Cauchy's theorem, $\int\limits_{\gamma} \frac{z}{(z1)(z2)} = 0$ when $2 < r < \infty$ because the integral is holomorphic inside the contour, and that $z > 2$ so since the values of $z$ are outside of the path, the integrand is holomorphic? The picture included with the solution is a circle on the complex plane centered at the origin with radius $2$ and everything outside the contour is shaded. I am confused about this; can someone clarify please?
