Cauchy-Goursat Theorem

**lacy1104** · April 14th 2008, 08:51 AM

Prove the following by means of Cauchy-Goursat theorem. Begin with

performed around |z|=1. Use the parametric representation

Separate your equation into real and imaginary parts.
12.

**ThePerfectHacker** · April 14th 2008, 08:59 AM

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 $\oint_{\Gamma} e^z dz = 0$ .

You prove this by writing,

$0=\oint \limits_{|z| = 1} e^z dz = \int_0^{2\pi} e^{e^{i\theta}} ie^{i\theta} d\theta \implies \int_0^{2\pi} e^{e^{i\theta}} e^{i\theta} = 0$

Now split the real and imaginary parts.
Hint: $e^{e^{i\theta}} = e^{\cos \theta} \cos (\sin \theta) + i e^{\cos \theta}\sin (\sin \theta)$ .

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