# Thread: f and conjugate are holomorphic

1. ## f and conjugate are holomorphic

I have to show that: if $f$ and $\overline{f}$ are holomorphic on an open connected set $U$, then $f$ is identically constant.

We can write $f= u(x,y)+iv(x,y)$
With the Cauchy-Riemann equations is easily shown that $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}=0$

However to show that $f$ is identically constant I don't see where I need to use that $U$ is a open set.

I understand that the condition "connected" is strictly necessary, otherwise f may take on different constant values on different components.

But does U necessarily need to be an open set. Is "containing an open ball" not enough?

2. Originally Posted by Dinkydoe
I have to show that: if $f$ and $\overline{f}$ are holomorphic on an open connected set $U$, then $f$ is identically constant.

We can write $f= u(x,y)+iv(x,y)$
With the Cauchy-Riemann equations is easily shown that $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}=0$

However to show that $f$ is identically constant I don't see where I need to use that $U$ is a open set.

I understand that the condition "connected" is strictly necessary, otherwise f may take on different constant values on different components.

But does U necessarily need to be an open set. Is "containing an open ball" not enough?
Requiring it to be open has more to do with the fact that a function $f:U\rightarrow \mathbb{C}$ is holomorphic at $z$ if there exists an open ball around $z$ such that $f$ is complex-differentiable there. So by the very definition, for $f$ to be holomorphic in $U$ each point of $U$ must have a neighbourhood contained in $U$ (too much $U$!).