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Math Help - Is the conjugate of a complex function holomorphic?

  1. #1
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    Is the conjugate of a complex function holomorphic?

    Hello

    Suppose f(z) is holomorphic on the open disc D(0,1) of the complex plane.

    How would one go about proving  g(z) = \overline{ f( \overline{z} ) } is holomorphic in D(0,1)?

    I attepted this by writing out the definition of the derivative, substituting \overline{ f( \overline{z} ) } = u(Re(z), - Im(z)) - i v(Re(z), - Im(z)) doing the algebra and using the facts that the real and imaginery parts of f are differentiable and if z \in D(0,1) then \overline{z}  \in D(0,1)

    Is this the way this question would be tackled?? i'm not sure if my proof is a hundred percent correct.

    Also I want to prove k(z)=\overline{ f( z ) } is differentiable at a \in D(0,1) iff f'(a) = 0.

    I'm a bit stuck on this part. Any ideas?

    Incidently this is question 5.11 from Introduction to Complex Analysis by Priestly.
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  2. #2
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    You would need to check the Cauchy-Riemann equations.
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    how can i check the cauchy conditions? i have no information about the continuity of the partial derivatives of the real and imaginary parts of f. I only know that f is holomorphic on the open disc and thus continuous but the cauchy equations only provide a sufficient condition for differentiability when the partial derivatives are continuous.
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  4. #4
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    First of all, when you're asking "Is the complex conjugate of a complex function holomorphic?" do you mean \displaystyle \overline{f(z)}?
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  5. #5
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    no i meant the function g(z) as defined above, didnt quite know how to word this briefly and accurately for the title.
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  6. #6
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    In general if U\subset \mathbb{C} is an open set and f:U\rightarrow \mathbb{C} is holomorphic then g(z)=\overline{f(\overline{z})} is holomorphic in U'=\{ z\in \mathbb{C} : \overline{z}\in U\}, and to see this just notice \frac{g(z)-g(z_0)}{z-z_0}=\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0}=\overline{ \left( \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } \rightarrow \overline{f'(\overline{z_0})}.

    For the second one use the Cauchy-Riemann equations together with the definition of the conjugation operation to conclude that the partials must be zero.
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  7. #7
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    Can you descirbe how the limit operation is performed please. I dont see how to take the limit of the lhs. Is this correct:

    \lim_{z \to z_0}<br />
\frac{g(z)-g(z_0)}{z-z_0}=\lim_{z \to z_0}\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0}=\lim_{z \to z_0}\overline{ \left( \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } = \overline{ \left( \lim_{z \to z_0}\frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } = \overline{f'(\overline{z_0})}

    I've used the fact z \mapsto \overline{z} is a continuous function so I can bring the limit inside the bracket. Is that right?
    Last edited by aukie; November 19th 2010 at 07:32 AM.
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