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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, 2013 at 16:14
  • $\begingroup$ Im guessing the path is something like $\gamma (t)=e^{it}$ with $0\leq t \leq 2\pi$ $\endgroup$
    – JamesT
    Apr 3, 2013 at 16:14
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    $\begingroup$ Have you tried that out? $\endgroup$
    – Clayton
    Apr 3, 2013 at 16:15
  • $\begingroup$ Yeah but the i get stuck with an $e^{e^{2it}}$ term and im not sure where to go then? $\endgroup$
    – JamesT
    Apr 3, 2013 at 16:16
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    $\begingroup$ 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, 2013 at 16:28

2 Answers 2

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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 $$\frac{1}{z}=\overline{z}$$

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  • $\begingroup$ Thats very helpful, taking a look a holomorphic functions now and it all makes sense $\endgroup$
    – JamesT
    Apr 3, 2013 at 16:34
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Hint: Set $z=e^{i\theta}$ for $0\leq\theta<2\pi$ and notice $dz=ie^{i\theta}d\theta$.

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