 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? Tags $gammatheta,$intlimitsgamma, fraczz1z2$, reitheta$ Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mared Geometry 1 June 15th, 2014 09:49 AM mauro125 Algebra 3 February 22nd, 2014 04:57 PM Jhenrique Calculus 1 November 23rd, 2013 10:02 AM kerrymaid Algebra 3 June 24th, 2010 10:37 AM FelisCanisOfCadog Algebra 4 March 6th, 2009 08:05 PM

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