Thread: f and conjugate are holomorphic

I have to show that: if and are holomorphic on an open connected set , then is identically constant.

We can write
With the Cauchy-Riemann equations is easily shown that

However to show that is identically constant I don't see where I need to use that 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?

Originally Posted by Dinkydoe

I have to show that: if and are holomorphic on an open connected set , then is identically constant.

We can write
With the Cauchy-Riemann equations is easily shown that

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