My Math Forum $\int\limits_{\gamma} \frac{z}{(z-1)(z-2)}$, $\gamma(\theta) = re^{i\theta}$

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 April 21st, 2015, 04:15 PM #1 Newbie   Joined: Apr 2015 From: Switzerland Posts: 1 Thanks: 0 $\int\limits_{\gamma} \frac{z}{(z-1)(z-2)}$, $\gamma(\theta) = re^{i\theta}$ $\int\limits_{\gamma} \frac{z}{(z-1)(z-2)}$, $\gamma(\theta) = re^{i\theta}$, $2 < r < \infty$ 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, $\int\limits_{\gamma} \frac{z}{(z-1)(z-2)} = 0$ 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?

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