Proof of maximum principle of Cauchy integral. I'm trying to proof the maximum principle, using the Cauchy integral, following this steps, but i'm stucked.
Maximum principle
If $f$ is a holomorphic function and not constant in a domain $D$, so $\left | f \right |$ cant't take any maximum in $D$.
With $f$ holomorphic function in a domain $D$.

*

*$a)$ First of all starting with the Cauchy's integral, we find the following identity:
$$
f^{k}(z_0)=\frac{k!}{2 \pi i}\oint \frac{f(z)dz}{(z-z_0)^{k+1}}
$$
Taking $(k=0)$ and $z=z_0+re^{i\theta} \quad \Rightarrow \quad dz=ire^{i\theta}d\theta$, we get:
$$
f(z_0)=\frac{1}{2\pi i}\oint \frac{f(z)dz}{(z-z_0)}\Rightarrow f(z-re^{i\theta})=\frac{1}{2\pi i}\int \frac{f(z)ire^{i\theta}}{re^{i\theta}}d\theta\Rightarrow \left \{ z \rightarrow z+re^{i\theta}\right \} \Rightarrow f(z)=\frac{1}{2\pi}\int_0^{2\pi} f(z+re^{i\theta})d\theta
$$


*$b)$ From this, instantly we get the expression:
$$
f(z)rdr=\frac{1}{2\pi}\int_0^{2\pi} f(z+re^{i\theta})d\theta rdr \quad \Rightarrow \quad f(z)= \frac{1}{\pi R^2}\int_0^{2\pi}\int_0^Rf(z+re^{i\theta})d\theta rdr
$$
That means the value of $f(z)$ is the average of the values of $f$ in a disk of radium $R$.


*$c)$ Proof that $\left | f\right |$ is a constant in the previous disk. We supose that exists a neighbourhood of $z$ $ \forall \zeta$ that $\left | f(\zeta)\right |\leq \left | f(z)\right |$, in a disk in this neighbourhood.
(I'm not sure how to follow from this step..)


*$d)$ Proof that if $\left | f\right |$ is a constant $f$ is too a constant in the same disk.


*$e)$ The previous steps proof the principle in a disk centered in $z$. Valid $\forall z$, so this is prooves in all the domain.
I need some help in the last steps. Thanks!
 A: Suppose $r\in \Bbb R^+$  and $D\supset B(z_0,r) $ and $|f(z_0)|=\max \{|f(z): z\in B(z_0,r)\}.$ Then for any $s\in (0,r)$ we have $$|f(z_0)|=\left| \frac {1}{2\pi}\int_{|z-z_0|=s}\frac {f(z)}{z-z_0}dz\right|    \le  \frac {1}{2\pi} \int_{|z-z_0|=s}\frac {|f(z)|}{|z-z_0|}|dz|\le$$ $$\le 
\frac {1}{2\pi}\int_{|z-z_0|=s}\frac {|f(z_0)|}{r}|dz|=|f(z_0)|.$$  So the 1st & 2nd "$\le$" above are actually "$=$". But since $f$ is continuous, if $|z_1-z_0|=s$ and $|f(z_1)|<|f(z_0)|$ then the 2nd "$\le$" above would be "$<$".
So $\forall z\in B(z_0,r)\,(|f(z)|=|f(z_0)|).$
Let $g=Re(f)$ and $h=Im(f).$ Let the subscripts $_1$ and $_2$ denote partial derivatives with respect to $Re(z)$ and $Im(z).$ At any $z\in B(z_0,r)$ we have $$0=(g^2+h^2)_1=2(gg_1+hh_1)$$ and $$0=(g^2+h^2)_2=2(gg_2+hh_2)=2(-gh_1+hg_1).$$ Now for  real $g,h,g_1,h_1,g_2,h_2$ we cannot have $gg_1+hh_1=0=-gh_1+hg_1$  unless $g=h=0$ or $g_1=h_1=0.$
(i). If $g=h=0$ at even one $z\in B(z_0,r)$ then $|f(z_0)|=|f(z)|=0$ so $f$ is constantly $0$ in $B(z_0,r).$
(ii). If $(g\ne 0\lor h\ne 0)$ at all $z\in B(z_0,r)$ then $\forall z\in B(z_0,r)\,(f'(z)=g_1(z)+ih_1(z)=0)$ so $f$ is constant in $B(z_0,r).$
DIGRESSION:We can use the Maximum Principle to prove the Fundamental Theorem of Algebra (Gauss):  If $p$ is a polynomial on $\Bbb C$ and $\forall z\in \Bbb C\,(p(z)\ne 0)$ then $p$ is constant. Proof:   Suppose $p$ is not constant. Then $|p(z)|\to \infty$ as $|z|\to \infty,$ so take $A\in \Bbb R^+$ such that $|z|>A\implies |p(z)|>|p(0)|.$ Now $\min \{|p(z)|:|z|\le A\}$ exists because $z\to |p(z)|$ is continuous. So if $p(z)$ is never $0$ then the function $q(z)=1/p(z)$ is holomorphic on $\Bbb C,$ and $\max \{|q(z)|: z\in \Bbb C\}=\max \{|q(z)|: |z|\le A\}$ exists , so $q,$ and hence $p$ also, are constant,... a contradiction.
