I have encountered a pretty classic statement, but in the book I am working through it is presented prior to the Cauchy Integral Theorem. Here is the exercise that I'm sure most of you are familiar with:
Let $f$ be analytic about $z_0$ with radius of convergence $R$, then for $0 < r < R$ we have $f(z_0) = \int_0^{2\pi} f(z_0 + re^{it})\frac{dt}{2\pi}$
Then use this to prove the Maximum Modulus principle.
My idea for the first part is to just use the power series centered at $z_0$, say $f(z) = \sum_0^{\infty}a_n(z-z_0)^n$, where $f(z_0) = a_0$. Then $f(z_0+re^{it}) = \sum_0^{\infty}a_n((z_0+re^{it})-z_0)^n = \sum_0^{\infty}a_nr^n(e^{int})$.
Then since we're in the disc of radius $R$ about $z_0$ away from the boundary, our series converges uniformly on the circle of radius $r$, so it should follow that we can pass the integral under the sum.
Thus $\int_0^{2\pi} f(z_0 + re^{it})\frac{dt}{2\pi} = \int_0^{2\pi}\sum_0^{\infty}a_nr^n(e^{int})\frac{dt}{2\pi} = \sum_0^{\infty}a_nr^n\int_0^{2\pi}(e^{int})\frac{dt}{2\pi}$.
This should give us the first part, but I have no idea how to turn this into the maximum modulus! Can anyone help?
Thanks a lot in advance!