Prove that the integral of $ exp(i heta) $ over a complete unit circle is zero.

To prove that the integral of $ \exp(i \theta) $ over a complete unit circle is zero, we evaluate the contour integral:

$ \int_0^{2\pi} \exp(i \theta) d\theta $

Recall that $ \exp(i \theta) = \cos(\theta) + i \sin(\theta) $. So the integral becomes:

$ \int_0^{2\pi} \cos(\theta) d\theta + i \int_0^{2\pi} \sin(\theta) d\theta $

We know that the integrals of $ \cos(\theta) $ and $ \sin(\theta) $ over a complete period from $ 0 $ to $ 2 \pi $ are both zero:

$ \int_0^{2\pi} \cos(\theta) d\theta = 0 $

$ \int_0^{2\pi} \sin(\theta) d\theta = 0 $

Thus, the original integral evaluates to:

$ \int_0^{2\pi} \exp(i \theta) d\theta = 0 $

To prove that the integral of $ exp(i heta) $ over a complete unit circle is zero, we start with:

$ int_0^{2pi} exp(i heta) d heta $

Using Euler

Evaluate:

$ int_0^{2pi} exp(i heta) d heta $

Since:

$ int_0^{2pi} cos( heta) d heta = 0 $

And:

$ int_0^{2pi} sin( heta) d heta = 0 $

Thus:

$ int_0^{2pi} exp(i heta) d heta = 0 $

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