Evaluate the integral
$\int_{\gamma}e^{z^2}+ \overline{z} \ \ dz, $
where $\gamma$ is the positively oriented unit circle.
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$\begingroup$ What steps have you taken to solve the problem? $\endgroup$ – Ron Gordon Apr 3 '13 at 16:14
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$\begingroup$ Im guessing the path is something like $\gamma (t)=e^{it}$ with $0\leq t \leq 2\pi$ $\endgroup$ – JamesT Apr 3 '13 at 16:14
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1$\begingroup$ Have you tried that out? $\endgroup$ – Clayton Apr 3 '13 at 16:15
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$\begingroup$ Yeah but the i get stuck with an $e^{e^{2it}}$ term and im not sure where to go then? $\endgroup$ – JamesT Apr 3 '13 at 16:16
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1$\begingroup$ Let $\gamma$ be a closed loop. $\gamma$ bounds a region $U$. If on a domain $D\supset \gamma\cup U$, $f$ is analytic, then$\int _{\gamma} f(z) dz=0$. Do you know this result?# $\endgroup$ – Spook Apr 3 '13 at 16:28
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Hint: $$\int_\gamma e^{z^2} + \overline{z} \; \mathrm{d}z= \int_\gamma e^{z^2} \; \mathrm{dz}+\int_{\gamma} \overline{z} \; \mathrm{d}z$$ For the first one use that $e^{z^2}$ is holomorphic (so what is the value of the integral)? for the second use a parametrisation of the unit circle.
Just cause i like it, you can use that for $|z|=1$ the following equlity holds $$\frac{1}{z}=\overline{z}$$
answered Apr 3 '13 at 16:27
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$\begingroup$ Thats very helpful, taking a look a holomorphic functions now and it all makes sense $\endgroup$ – JamesT Apr 3 '13 at 16:34
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Hint: Set $z=e^{i\theta}$ for $0\leq\theta<2\pi$ and notice $dz=ie^{i\theta}d\theta$.
