# Thread: f and conjugate are holomorphic

1. ## f and conjugate are holomorphic

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

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

However to show that $\displaystyle f$ is identically constant I don't see where I need to use that $\displaystyle U$ 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 $\displaystyle f$ and $\displaystyle \overline{f}$ are holomorphic on an open connected set $\displaystyle U$, then $\displaystyle f$ is identically constant.

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

However to show that $\displaystyle f$ is identically constant I don't see where I need to use that $\displaystyle U$ 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 $\displaystyle f:U\rightarrow \mathbb{C}$ is holomorphic at $\displaystyle z$ if there exists an open ball around $\displaystyle z$ such that $\displaystyle f$ is complex-differentiable there. So by the very definition, for $\displaystyle f$ to be holomorphic in $\displaystyle U$ each point of $\displaystyle U$ must have a neighbourhood contained in $\displaystyle U$ (too much $\displaystyle U$!).