If $f(z)$ is analytic, and $\overline{f(z)}$ is analytic, then is $f$ necessarily a constant function?

Question

I need help with verifying my following proof. It feels a little fishy to me.

If $f(z)$ is analytic, and $\overline{f(z)}$ is analytic, then is $f$ necessarily a constant function?

We know $f(z)=u(x,y)+iv(x,y)$ and $\overline{f(z)}=u(x,y)+iv'(x,y)$, where $v'=-v$. $f$ satisifies the Cauchy Riemann equations, thus,

For $f$, one has that: $u_x=v_y, v_x=-u_y$.

For $\overline{f}$, one has that:

$u_x=v'_y=-v_y$

$v'_x=-v_x=--u_y$.

One has $u_x=-v_y=v_y$, which forcefully makes $v_y=0$. Also, $u_y=v_x=-v_x$, so $v_x=0$. So for all $z$, $f'(z)=0$ and this shows that $f$ is a constant function.

Does this proof work?

Answer 1

Your proof is correct. You can write it more compactly as

• $f=u+iv$ is analytic $\implies u_x=v_y,u_y=-v_x$,
• $\overline f=u-iv$ is analytic $\implies u_x=-v_y,u_y=v_x$,

and all derivatives must be zero.

Answer 2

Your proof is correct. My proof: let $g(z):=f(z)\overline{f(z)}.$ Then $g(z)=|f(z)|^2$ is analytic and real(!). Conclusion ?

Answer 3

Here is an alternative:

$f+\overline f=\Re(f)$ is analytic. Now, the image of $\Re(f)$ is a subset of $\mathbb R$ hence not open in $\mathbb C$. It follows by open mapping theorem that $\Re (f)$ must be a constant. Similarly, $\Im (f)$ is a constant. Therefore, $f=\Re(f)+\Im(f)$ is a constant.