# Complex Analysis/Application of Cauchy-Riemann Equations
## Statement

It is G⊆C{\textstyle G\subseteq \mathbb {C} } an area, f:G→C{\textstyle f\colon G\to \mathbb {C} } holomorphic. If |f|{\textstyle |f|} is constant to G{\textstyle G}, then f{\textstyle f} is constant.

## Proof

It is open G⊆C{\textstyle G\subseteq \mathbb {C} }, f:G→C{\textstyle f\colon G\to \mathbb {C} } holomorphic. Other |f|{\textstyle |f|} constant.

### Proof of Lemmas 1

If |f|=√(Re(f)$^2$+Im(f)$^2$){\textstyle |f|={\sqrt {{\mathfrak {Re}}(f)^{2}+{\mathfrak {Im}}(f)^{2}}}} is constant to D$_r$(z$_0$){\textstyle D_{r}(z_{0})}, then also Re(f)$^2$+Im(f)$^2$=c{\textstyle {\mathfrak {Re}}(f)^{2}+{\mathfrak {Im}}(f)^{2}=c} must be constant with a constant c∈R{\textstyle c\in \mathbb {R} }. If h:=Re(f)$^2$+Im(f)$^2${\textstyle h:={\mathfrak {Re}}(f)^{2}+{\mathfrak {Im}}(f)^{2}} is constant, the partial derivation (∂h(x+iy))/∂x=0{\textstyle {\frac {\partial h(x+iy)}{\partial x}}=0} and (∂h(x+iy))/∂y=0{\textstyle {\frac {\partial h(x+iy)}{\partial y}}=0}.

### Proof of the Lemmas 2

Because of f{\textstyle f} holomorphic on D$_r$(z$_0$){\textstyle D_{r}(z_{0})} the Cauchy-Riemannschen equations apply to
    $u:={\mathfrak {Re}}(f),\ v:={\mathfrak {Im}}(f)\ {\mbox{ und }}\ f(z):=u(x,y)+i\cdot v(x,y)$
and
    ${\frac {\partial {\bigu(x,y)^{2}+v(x,y)^{2}{\big )}}{\partial x}}=0{\mbox{ und }}{\frac {\partial {\bigu(x,y)^{2}+v(x,y)^{2}{\big )}}{\partial y}}=0$

### Proof of the Lemmas 3

If u$_x$=∂u/∂x {\textstyle u_{x}={\frac {\partial u}{\partial x}}} and u$_y$=∂u/∂y {\textstyle u_{y}={\frac {\partial u}{\partial y}}} and application of the chain rule to the partial derivations are obtained the two equations
    $0=2uu_{x}+2vv_{x}$ and 0=2uu$_y$+2vv$_y${\textstyle 0=2uu_{y}+2vv_{y}}
With CR-DGL u$_x$=v$_y${\textstyle u_{x}=v_{y}} and u$_y$=−v$_x${\textstyle u_{y}=-v_{x}}, the partial derivation of v{\textstyle v} is replaced by partial derivations of u{\textstyle u} and obtained (factor 2 can be omitted):
    $0=2\cdot (uu_{x}-vu_{x})$ and 0=2⋅(uu$_y$+vu$_x$){\textstyle 0=2\cdot (uu_{y}+vu_{x})}

### Proof of the Lemmas 4

We square the two equations
    $0=(uu_{x}-vu_{y})^{2}=u^{2}u_{x}^{2}-2uu_{x}\cdot vu_{y}+v^{2}u_{y}^{2}$
    $0=(uu_{y}+vu_{x})^{2}=u^{2}u_{y}^{2}+2uu_{y}\cdot vu_{x}+v^{2}u_{x}^{2}$
and add these two squared equations to:
    $0=u^{2}u_{x}^{2}+v^{2}u_{y}^{2}+u^{2}u_{y}^{2}+v^{2}u_{x}^{2}$

### Proof of the Lemmas 5

Clamping u$^2${\textstyle u^{2}} and v$^2${\textstyle v^{2}} gives:
    $0=u^{2}(u_{x}^{2}+u_{y}^{2})+v^{2}(u_{x}^{2}+u_{y}^{2})=(u^{2}+v^{2})\cdot (u_{x}^{2}+u_{y}^{2})$
This follows with the real-value component or Imaginary part functions in the product:
    $0=u^{2}+v^{2}\vee 0=u_{x}^{2}+u_{y}^{2}$

### Proof of the Lemmas 6

* $u$ and $v$ are real-valued and with u$^2$=−v$^2${\textstyle u^{2}=-v^{2}} the only option to fulfill the equation is u=v=0{\textstyle u=v=0} i.e. u(x, y)=0{\textstyle u(x, y)=0} and v(x, y)=0{\textstyle v(x, y)=0}. This implies that $f$ is constant with $f=u+i\cdot v=0$.
* Similar to the argument above u$_x$$^2$=−u$_y$$^2${\textstyle u_{x}^{2}=-u_{y}^{2}} implies u$_x$$^2$=u$_y$$^2$=0{\textstyle u_{x}^{2}=u_{y}^{2}=0} for the partial derivatives and u$_x$=u$_y$=0{\textstyle u_{x}=u_{y}=0}. With the application of the Cauchy-Riemann Equations and $f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)=0$
    for $z=x+iy$.
In both cases f{\textstyle f} is constant on G{\textstyle G}.