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

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In summary, a contour integral is an integral evaluated along a specific path in the complex plane. To calculate it, a path is chosen and parameterized, and the fundamental theorem of calculus is used. The function e^{i \theta} is significant in contour integrals as it simplifies the calculation process. The interval of integration is typically from 0 to 2\pi, corresponding to a full rotation around the unit circle in the complex plane. If the contour integral equals 0, it means the area under the curve is also 0, with varying interpretations depending on the context and function.
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Homework Statement


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


Homework Equations



Cauchy's theorem: [tex]\oint f(z)dz=0[/tex]

The Attempt at a Solution



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

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

What is a contour integral?

A contour integral is an integral that is evaluated along a specific path in the complex plane, rather than a straight line. It is used to calculate the area under a curve in the complex plane.

How do you calculate a contour integral?

To calculate a contour integral, you need to choose a path in the complex plane, parameterize it using a variable such as t, and then use the fundamental theorem of calculus to evaluate the integral along that path.

What is the significance of the function e^{i \theta} in the given integral?

The function e^{i \theta} is a complex exponential function that represents a point on the unit circle in the complex plane. It is often used in contour integrals because it simplifies the calculation process and allows for easier evaluation of the integral.

Why is the interval of integration from 0 to 2\pi?

The interval of integration from 0 to 2\pi corresponds to a full rotation around the unit circle in the complex plane. This allows for the integral to capture all possible values of the function along the chosen path.

What does it mean for the contour integral to equal 0?

If the contour integral is equal to 0, it means that the area under the curve in the complex plane is equal to 0. This can have various interpretations, depending on the context of the problem and the function being integrated.

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