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I have $$\int_{\gamma}\frac{1}{4z^2-1}dz$$, where $\gamma$ is the unit circle in the complex plane.

I said this integral equals to $$\int_{0}^{2\pi}\frac{ie^{it}}{4(e^{it})^2-1}dt$$ Then I let $u=e^{it}$

which gave me $$\int_{1}^{1}\frac{du}{4u^2-1}$$ which should equal 0.. but is this possible because there are roots -1/2 and +1/2 inside of the closed contour $\gamma$ so how can it be 0 if it's not satisfying the criteria for Cauchy's Theorem?

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  • $\begingroup$ When you attempt the substitution $u=e^{it}$ you are effectively taking the $\log$ which is not analytic on the entire unit circle... $\endgroup$ – copper.hat Mar 3 '14 at 8:04
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It might be easier to expand the integrand as $\frac{1}{4z^2-1} = {1 \over 4} \left( \frac{1}{z-{1 \over 2} }-\frac{1}{z+{1 \over 2} } \right)$. (Hence the integral is zero.)

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  • $\begingroup$ Ah we can just use the Cauchy Integral formula to get $\frac{1}{4}2\pi i (1) - \frac{1}{4}2\pi i (1)$ ? $\endgroup$ – user84226 Mar 3 '14 at 8:20
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    $\begingroup$ Sure, use $f(z) = 1$ in $f(a) = {1 \over 2 \pi i} \int { f(z) \over z-a} dz$. $\endgroup$ – copper.hat Mar 3 '14 at 8:22

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