I am asked to show that $$|f(z)| \le max_{\gamma} |f(z+re^{it})|$$ and I've done that proof. I'm stuck on proving that $|f(z)|$ has no local maxima (within its domain of analyticity) or minima (assuming f is analytic and never zero on it's domain).
For reference, this pertains to the Cauchy Integral Formula, particularly the case where $z$ is a point in the center of the circle $\gamma$. This case yields: $$f(z) = \frac{1}{2 \pi} \int_{0}^{2\pi} f(z+re^{it})dt$$
I'm just at a loss for a way to put all this together. I'm not asking for the answer, just some place to start.
Well, suppose $|f(z)|$ has a maximum at $z_0$ - that means the value at $z_0$ is strictly bigger than any value of $|f(z)|$ on any small neighborhood of $z_0$, and in particular a small circle around $z_0$. But this contradicts the inequality you began your question with. For local minimum, apply the same argument to $1/f$.