$f$ is constant complex function if $Af(z) +B \overline{f(z)} = 0$ for some $A$ and $B \in \mathbb{C}$

Question

Let $U \subset \mathbb{C}$ be a open convex set and $f : U \to C$ a analytic function, such that there exist $A, B \in \mathbb{C}$ and

$ Af(z) +B\overline{f(z)}=0$

Show that $f(z)$ is constant.

I try the follow if $A$ or $B$ ais zero ready!!

if $A$ and $B$ are not zero

$ Af(z) +B\overline{f(z)}=0$ then $f^2(z) = c|f(z)|^2$ so $f(z) \in c\cdot \mathbb{R}$ so $f^2(z)$ is not open.

I would like to prove that f 'is zero some help?.

thank you

Answer 1

If $f$ is a holomorphic function and $f'(z_0)\neq0$, then $\overline f$ is not differentiable at $z_0$. In fact, if $f(x+yi)=u(x,y)+v(x,y)i$, and if $z_0=x_0+y_oi$, then $\frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0)$ and $\frac{\partial u}{\partial y}(x_0,y_0)=-\frac{\partial v}{\partial x}(x_0,y_0)$ (the Cauchy-Riemann equations). But $\overline{f(x+yi)}=u(x,y)-v(x,y)i$, and therefore you can't also have $\frac{\partial u}{\partial x}(x_0,y_0)=-\frac{\partial v}{\partial y}(x_0,y_0)$ and $\frac{\partial u}{\partial y}(x_0,y_0)=\frac{\partial v}{\partial x}(x_0,y_0)$, unless all these numbers are $0$.

So (assuming that $A,B\neq0$) if $Af+b\overline f=0$, then $\overline f=-\frac ABf$ and therefore $f$ and $\overline f$ are both holomorphic. It follows from what I proved above that $f'$ is zero everywhere. Therefore, $f$ is constant.

Answer 2

If $(A,B)\ne(0,0)$ but $A=0$ or $B=0$, then $f\equiv0$.

If $A,B\ne 0$, $c=A/B$ and $f=u+iv$, then $cf+\overline f=0$, and hence $$ cf^2=-\overline{f}f=-|f|^2 $$ hence the analytic function $g=cf^2$ takes only real values, and $g$ is constant. Thus $f$ is also constant.