proof of Gauss’ mean value theorem

We can parameterize the circle by letting $z=z_0+r{e}^{i\varphi }$. Then $dz=ir{e}^{i\varphi }d\varphi$. Using the Cauchy integral formula we can express $f(z_0)$ in the following way:

$f(z_0)$ $=$ ${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\oint }_C}{\displaystyle \frac{f(z)}{z-z_0}}𝑑z$

$=$ ${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_0^{2\pi }}{\displaystyle \frac{f(z_0+r{e}^{i\varphi })}{r{e}^{i\varphi }}}ir{e}^{i\varphi }𝑑\varphi$

$=$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}f(z_0+r{e}^{i\varphi })𝑑\varphi .$

Title	proof of Gauss’ mean value theorem
Canonical name	ProofOfGaussMeanValueTheorem
Date of creation	2013-03-22 13:35:36
Last modified on	2013-03-22 13:35:36
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	18
Author	yark (2760)
Entry type	Proof
Classification	msc 30E20
Related topic	CauchyIntegralFormula