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

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?

• The proof looks good. – Paul K Jan 14 '20 at 7:53

## 2 Answers

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.

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 ?

• It is more "My beginning of a proof" ;) – nicomezi Jan 14 '20 at 10:08