Complex Analysis/Application of Cauchy-Riemann Equations

Statement

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

Proof

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

Proof of Lemmas 1

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

Proof of the Lemmas 2

Because of ${\textstyle f}$ holomorphic on ${\textstyle D_{r}(z_{0})}$ 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 {\big (}u(x,y)^{2}+v(x,y)^{2}{\big )}}{\partial x}}=0{\mbox{ und }}{\frac {\partial {\big (}u(x,y)^{2}+v(x,y)^{2}{\big )}}{\partial y}}=0

Proof of the Lemmas 3

If ${\textstyle u_{x}={\frac {\partial u}{\partial x}}}$ and ${\textstyle u_{y}={\frac {\partial u}{\partial y}}}$ and application of the chain rule to the partial derivations are obtained the two equations

0=2uu_{x}+2vv_{x}

and

{\textstyle 0=2uu_{y}+2vv_{y}}

With CR-DGL ${\textstyle u_{x}=v_{y}}$ and ${\textstyle u_{y}=-v_{x}}$ , the partial derivation of ${\textstyle v}$ is replaced by partial derivations of ${\textstyle u}$ and obtained (factor 2 can be omitted):

0=2\cdot (uu_{x}-vu_{x})

and

{\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 ${\textstyle u^{2}}$ and ${\textstyle v^{2}}$ 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 ${\textstyle u^{2}=-v^{2}}$ the only option to fulfill the equation is ${\textstyle u=v=0}$ i.e. ${\textstyle u(x,y)=0}$ and ${\textstyle v(x,y)=0}$ . This implies that $f$ is constant with $f=u+i\cdot v=0$ .
Similar to the argument above ${\textstyle u_{x}^{2}=-u_{y}^{2}}$ implies ${\textstyle u_{x}^{2}=u_{y}^{2}=0}$ for the partial derivatives and ${\textstyle u_{x}=u_{y}=0}$ . 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 ${\textstyle f}$ is constant on ${\textstyle G}$ .

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