A harmonic function cannot attain a strict local max in the domain 
I feel like I understand how the max principle works, but I don't understand the hint that was given on this problem. I am working on studying for an exam in a couple weeks. Any help would be much appreciated on how to use the hint.
The proof that I currently have is that by the mean value theorem, $$|f(a)|\leq |\frac1{2 \pi i}|\int_0^{2\pi} |f(a+Re^{i\theta})|d\theta$$
Suppose for the sake on contradiction that a is a strict local max in the domain D. Then $$|f(a)|> |\frac1{2 \pi i}|\int_0^{2\pi} |f(a+Re^{i\theta})|d\theta$$ which contradicts the above inequality. So we have that $|f(z)|$ cannot attain a strict local max in D. 
 A: Let $v(x, y)$ be a harmonic conjugate of $u(x, y)$ in $D$; then
$f(z) = f(x + iy) = u(x, y) + i v(x, y) \tag 1$
is holomorphic in $D$, as is well-known.  Thus, so
$g(z) = e^{f(z)} = e^{u(x, y) + i v(x, y)} \tag 2$
is also holomorphic in $D$; this implies, by the maximum modulus principle, that $\vert g(z) \vert$ cannot attain a strict local maximum in $D$; but
$\vert g(z) \vert = \vert e^{f(z)} \vert = \vert e^{u(x, y) + i v(x, y)} \vert = \vert e^{u(x, y)} e^{i v(x, y)} \vert = \vert e^{u(x, y)} \vert \vert e^{i v(x, y)} \vert = e^{u(x, y)}, \tag 3$
since
$\vert e^{i v(x, y)} \vert = \vert \cos v(x, y) + i \sin v(x, y) \vert = 1 \tag 4$
and, since $e^{u(x, y)}$ is positive real,
$\vert e^{u(x, y)} \vert = e^{u(x, y)}; \tag 5$
since $\vert g(z) \vert$ cannot attain a strict local maximum in $D$, neither can the function $u(x, y)$, since it is, by (3), the log modulus of the holomorphic function $g(z)$:
$u(x, y) = \ln \vert g(z) \vert, \tag 6$
and $\ln$ is strictly monotonically increasing on the positive reals.
