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

[sinned4789]

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

 

[Dick]

Just put zeta=z+r*e^(i*t). It's a simple substitution problem.