if $f(z),\overline {f(z)}$ are analytic then they are constant

Question

I'm trying to prove this "theorem":

if $f(z),\overline {f(z)}$ are analytic in some open set $\Omega \subseteq \mathbb C$, then $f(z)$ is a constant.

Hint: Use Cauchy-Riemann equations to show that anywhere in $\Omega$: $\frac{\partial f}{\partial y}=\frac{\partial f}{\partial x}=0$

If $f$ and $\overline {f}$ are analytic in $\Omega$, then from CR we know that $\frac{\partial f}{\partial y}=i\frac{\partial f}{\partial x}$ and $\frac{\partial \overline f}{\partial y}=i\frac{\partial \overline f}{\partial x}$

But we can't really say if there is any connection between the derivatives. Notice that we are not talking about the derivative of $f$ with respect to $\overline z$, but rather, the derivative of $\overline f$!

I'd appreciate a hint in the right direction

Answer 1

Note that if $f$ and $\bar f$ are analytic, then $\Re f$ and $\Im f$ are analytic. But these take only real values.

Answer 2

Daniel Fischer's comment is one way: if $f$ and $\overline f$ are analytic, then so is $f\,\overline f = \|f\|^2$, and it's not hard to show that cannot satisfy the Cauchy–Riemann equations. A weakness in this argument is that it cannot be fully understood by someone who has not yet seen a proof that if $f$ and $g$ are analytic then is is $fg$.

Answer 3

Suppose we have a domain $E \subseteq \mathbb{C}$ and that $f$ is analytic. We can write $f(z)=f(x,y)=u(x,y)+iv(x,y)$ and we know, since $f$ is analytic, that

\begin{align*} u_x&=v_y\\ u_y&=-v_x \end{align*}

We now consider $\overline{f}(x,y)=u(x,y)-iv(x,y)$. We will get the negative of our derivatives above and we have

\begin{align*} u_x&=-v_y\\ u_y&=v_x \end{align*}

This is only true if $u_x=u_y=v_x=v_y=0$ and so $f$ must be constant.