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Math Help - f and conjugate are holomorphic

  1. #1
    Senior Member Dinkydoe's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by Dinkydoe View Post
    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!).
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