1
$\begingroup$

Evaluate the integral

$\int_{\gamma}e^{z^2}+ \overline{z} \ \ dz, $

where $\gamma$ is the positively oriented unit circle.

$\endgroup$
9
  • $\begingroup$ What steps have you taken to solve the problem? $\endgroup$ – Ron Gordon Apr 3 '13 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 '13 at 16:14
  • 1
    $\begingroup$ Have you tried that out? $\endgroup$ – Clayton Apr 3 '13 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 '13 at 16:16
  • 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
3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thats very helpful, taking a look a holomorphic functions now and it all makes sense $\endgroup$ – JamesT Apr 3 '13 at 16:34
2
$\begingroup$

Hint: Set $z=e^{i\theta}$ for $0\leq\theta<2\pi$ and notice $dz=ie^{i\theta}d\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.