Let a be the average value of |f| on a small circle with center c. Rewrite f(c) using the Cauchy integral formula. Thus 2πi×f(c) equals the contour integral of f(z)/(z-c), running around our small circle of radius r. Take norms across the board. The norm of z-c is r. And the norm of f(z) is no larger than a. The direction vector, running around the circle, has norm r. The r's cancel, and we have |f(c)| ≤ a.
Let m be the maximum of the continuous function |f| over the closed disk. Therefore |f(c)| ≤ a ≤ m.
If f attains a local maximum at c, then f(c) = m. Then means a = m, and |f(z)| = m on the circle of radius r. This holds for all smaller concentric circles, hence |f| is constant on a disk about c.
We know |f| is constant; does this mean f is constant?
Write f as u(x,y) + v(x,y)i. Start with u2+v2 = m and differentiate with respect to x, and with respect to y. Thus u times the partial of u with respect to x = -v times the partial of v with respect to x, and similarly for y.
Divide these equations to show u∂x over u∂y = v∂x over v∂y.
recall the Cauchy Riemann condition for analytic functions.
u∂x = v∂y
u∂y = -v∂x
Substitute in the previous equation to get (u∂x)2 = -(u∂y)2. If a square equals minus a square, then both numbers are 0. Thus the partials of u are zero, and a similar argument shows the partials of v are 0. Therefore u and v are constant, and f is constant.
If an analytic function is not constant, it is increasing in some direction.
As a corollary, f cannot attain a local minimum unless |f| = 0. Otherwise 1/f would attain a local maximum.
The function z2 has a minimum norm at 0, and |z2| is positive elsewhere.