Contour Integrals: Show \int_{0}^{2\pi} e^{i \theta}f(e^{i \theta}) d\theta=0

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SUMMARY

The integral \(\int_{0}^{2\pi} e^{i \theta}f(e^{i \theta}) d\theta\) equals zero when \(f(z)\) is analytic on and inside the unit circle \(|z|=1\). This conclusion is derived from Cauchy's theorem, which states that the integral of an analytic function over a closed contour is zero. By substituting \(z = e^{i\theta}\), the differential \(dz\) can be expressed as \(i e^{i\theta} d\theta\), allowing the transformation of the integral into a form that aligns with Cauchy's theorem.

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Homework Statement


If f(z) is analytic on and inside |z|=1, show \int_{0}^{2\pi} e^{i \theta}f(e^{i \theta}) d\theta = 0


Homework Equations



Cauchy's theorem: \oint f(z)dz=0

The Attempt at a Solution



I'm not really sure what to do here except setting z=e^{i\theta} and plugging in, but then i don't know what to do with the d\theta or how to make it look more like Cauchy's theorem :-\
 
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If z=e^(i*theta), what's dz?
 

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