Specializing Cauchy's Formula: Integral Limits 0-2\pi

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SUMMARY

This discussion focuses on specializing Cauchy's Integral Formula for the case where z is the center of the circle. The formula is expressed as f(z) = (1/2π)∫ f(z + re^(it)) dt, with integral limits from 0 to 2π. The original Cauchy's formula is f(z) = (1/i2π) ∫ f(ζ)/ (ξ - z) dζ. A key insight provided is the substitution ζ = z + r*e^(it), simplifying the problem significantly.

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Homework Statement


specialize cauchy's formula to the case when z is the center of the circle and show that

f(z) = (1/2\pi)\int f(z + re^(it)) dt
integral limits are 0 to 2\pi


Homework Equations


cauchy's formula
f(z) = (1/i2*pi) \int f(\zeta)/(\xi - z) d\zeta

it is 2*pi not 2^pi just to be sure


The Attempt at a Solution



i have no idea how to do this
any input or hints would be appreciated
 
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Just put zeta=z+r*e^(i*t). It's a simple substitution problem.
 
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