Show $f$ is constant given $g=\overline{f}$

Question

I am trying to show that if $f$ and $g=\overline{f}$ are both differentiable in a domain, then $f$ is constant on that domain.

My attempt: Let $$f(z)=u(x,y)+iv(x,y)$$ $$g(z)=r(x,y)+is(x,y)$$ Equating real and imaginary parts, we can see that $$u(x,y)=r(x,y) \ \ \ (1)$$ $$v(x,y)=-s(x,y) \ \ \ (2)$$ So we want to show that $u_x=u_y=v_x=v_y=0$. How can I show this?

I have tried differentiating $(1), (2)$ with respect to $x$ and $y$ to utilise the CR equations, but I have not shown the desired result.

Answer 1

The CR equations for $f$ give $u_x=v_y$; the CR equations for $g$ give $r_x=s_y$. Combining this with $u=r$ and $v=-s$ gives $u_x=v_y=-v_y$. Thus we must have $u_x=v_y=0$.

See if you can use a similar technique to derive $u_y=v_x=0$.

Answer 2

\begin{eqnarray*} f(x+\jmath y) &=& u(x,y) + \jmath v(x,y) \end{eqnarray*}

where $u(x,y)$ and $v(x,y)$ are real functions of two inputs $(x,y)\in \mathbb{R}^{2}$. The Cauchy-Riemann equations are, \begin{eqnarray*} \frac{\partial u}{\partial x} &=& \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &=& -\frac{\partial v}{\partial x} \end{eqnarray*}

SInce $ g\triangleq f^{\star}$,

\begin{eqnarray*} f(x+\jmath y) &=& u(x,y) + \jmath v(x,y) \\ g(x+\jmath y) &=& u(x,y) - \jmath v(x,y) \end{eqnarray*}

The CR then will takes us to, \begin{eqnarray*} \frac{\partial u}{\partial x} &=& \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &=& -\frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial x} &=& -\frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &=& \frac{\partial v}{\partial x} \\ \end{eqnarray*}

This is satisfied when \begin{eqnarray*} \frac{\partial u}{\partial x} &=& 0\\ \frac{\partial u}{\partial y} &=& 0\\ \frac{\partial v}{\partial x} &=&0 \\ \frac{\partial v}{\partial y} &=&0\\ \end{eqnarray*}

That is, $f$ (and $g$) are constants.