Editing Maximum modulus principle (section)

==Sketches of proofs==

===Using the maximum principle for harmonic functions===
One can use the equality

:<math>\log f(z) = \ln |f(z)| + i\arg f(z)</math>

for complex [[natural logarithm]]s to deduce that <math> \ln |f (z) | </math> is a [[harmonic function]]. Since <math>z_0</math> is a local maximum for this function also, it follows from the [[maximum principle]] that  <math>| f (z) | </math>  is constant. Then, using the [[Cauchy–Riemann equations]] we show that <math>f'(z)</math> = 0, and thus that <math>f(z)</math> is constant as well. Similar reasoning shows that  <math> | f (z) | </math>  can only have a local minimum (which necessarily has value 0) at an isolated zero of <math>f(z)</math>.

===Using Gauss's mean value theorem===
Another proof works by using [[Cauchy's integral formula#Consequences|Gauss's mean value theorem]] to "force" all points within overlapping open disks to assume the same value as the maximum. The disks are laid such that their centers form a polygonal path from the value where <math>f(z)</math> is maximized to any other point in the domain, while being totally contained within the domain. Thus the existence of a maximum value implies that all the values in the domain are the same, thus <math>f(z)</math> is constant.

===Using Cauchy's Integral Formula===
Source:<ref>{{cite book |last1=Conway |first1=John B. |editor1-last=Axler |editor1-first=S. |editor2-last=Gehring |editor2-first=F.W. |editor3-last=Ribet |editor3-first=K.A. |title=Functions of One Complex Variable I |date=1978 |publisher=Springer Science+Business Media, Inc. |location=New York |isbn=978-1-4612-6314-2 |edition=2}}</ref>

As <math>D</math> is open, there exists <math>\overline{B}(a,r)</math> (a closed ball centered at <math>a \in D</math> with radius <math>r>0</math>) such that <math>\overline{B}(a,r) \subset D</math>. We then define the boundary of the closed ball with positive orientation as <math>\gamma(t)=a+re^{it}, t \in [0,2\pi]</math>. Invoking Cauchy's integral formula, we obtain

:<math> 0 \leq \int_{0}^{2\pi} |f(a)|-| f(a+re^{it})| \,dt \leq 0 </math>

For all <math>t \in [0,2\pi]</math>, <math>| f(a) |-| f(a+re^{it}) | \geq 0</math>, so <math>| f(a)|=| f(a+re^{it}) |</math>. This also holds for all balls of radius less than <math>r</math> centered at <math>a</math>. Therefore, <math>f(z)=f(a)</math> for all <math>z \in \overline{B}(a,r)</math>.

Now consider the constant function <math>g(z)=f(a)</math> for all <math>z \in D</math>. Then one can construct a sequence of distinct points located in <math>\overline{B}(a,r)</math> where the holomorphic function <math>g-f</math> vanishes. As <math>\overline{B}(a,r)</math> is closed, the sequence converges to some point in <math>\overline{B}(a,r) \in D</math>. This means <math>f-g</math> vanishes everywhere in <math>D</math> which implies <math>f(z)=f(a)</math> for all <math>z \in D</math>.