# Cauchy-Goursat Theorem

• April 14th 2008, 08:51 AM
lacy1104
Cauchy-Goursat Theorem
Prove the following by means of Cauchy-Goursat theorem. Begin with http://answerboard.cramster.com/Answ...1657843372.gif performed around |z|=1. Use the parametric representation http://answerboard.cramster.com/Answ...1032846094.gif Separate your equation into real and imaginary parts.
You do not even need Cauchy closed curve theorem here. Just note that $\left( e^z \right)' = e^z$ which means if $\Gamma$ it a piecewise smooth closed curve then $\oint_{\Gamma} e^z dz = 0$.
$0=\oint \limits_{|z| = 1} e^z dz = \int_0^{2\pi} e^{e^{i\theta}} ie^{i\theta} d\theta \implies \int_0^{2\pi} e^{e^{i\theta}} e^{i\theta} = 0$
Hint: $e^{e^{i\theta}} = e^{\cos \theta} \cos (\sin \theta) + i e^{\cos \theta}\sin (\sin \theta)$.