Complex Analysis/Application of Cauchy-Riemann Equations

Statement

It is $G\subseteq ℂ$ an area, $f:G\to ℂ$ holomorphic. If $|f|$ is constant to $G$, then $f$ is constant.

Proof

It is open $G\subseteq ℂ$, $f:G\to ℂ$ holomorphic. Other $|f|$ constant.

Proof of Lemmas 1

If $|f|=\sqrt{\Re 𝔢(f{)}^2+\Im 𝔪(f{)}^2}$ is constant to $D_r(z_0)$, then also $\Re 𝔢(f{)}^2+\Im 𝔪(f{)}^2=c$ must be constant with a constant $c\in ℝ$. If $h:=\Re 𝔢(f{)}^2+\Im 𝔪(f{)}^2$ is constant, the partial derivation $\frac{\partial h(x+iy)}{\partial x}=0$ and $\frac{\partial h(x+iy)}{\partial y}=0$.

Proof of the Lemmas 2

Because of $f$ holomorphic on $D_r(z_0)$ the Cauchy-Riemannschen equations apply to

$$u:=\Re 𝔢(f),v:=\Im 𝔪(f)\text{ und }f(z):=u(x,y)+i\cdot v(x,y)$$

and

$$\frac{\partial (u(x,y{)}^2+v(x,y{)}^2)}{\partial x}=0\text{ und }\frac{\partial (u(x,y{)}^2+v(x,y{)}^2)}{\partial y}=0$$

Proof of the Lemmas 3

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

$$0=2u{u}_x+2v{v}_x$$ and $0=2u{u}_y+2v{v}_y$

With CR-DGL $u_x=v_y$ and $u_y=-v_x$, the partial derivation of $v$ is replaced by partial derivations of $u$ and obtained (factor 2 can be omitted):

$$0=2\cdot (u{u}_x-v{u}_x)$$ and $0=2\cdot (u{u}_y+v{u}_x)$

Proof of the Lemmas 4

We square the two equations

$$0=(u{u}_x-v{u}_y{)}^2=u^2{u}_x^2-2u{u}_x\cdot v{u}_y+v^2{u}_y^2$$

$$0=(u{u}_y+v{u}_x{)}^2=u^2{u}_y^2+2u{u}_y\cdot v{u}_x+v^2{u}_x^2$$

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 $u^2$ and $v^2$ 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 $u^2=-v^2$ the only option to fulfill the equation is $u=v=0$ i.e. $u(x,y)=0$ and $v(x,y)=0$. This implies that $f$ is constant with $f=u+i\cdot v=0$.
• Similar to the argument above $u_x^2=-u_y^2$ implies $u_x^2=u_y^2=0$ for the partial derivatives and $u_x=u_y=0$. With the application of the Cauchy-Riemann Equations and $$f^{\prime }\left(z\right)=\frac{\partial u}{\partial x}(x,y)-i\frac{\partial u}{\partial y}(x,y)=0$$

for $z=x+iy$.

In both cases $f$ is constant on $G$.

See also

• Cauchy-Riemannschen Equations (CRE)

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