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Math Help - Complex Analysis

  1. #1
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    Complex Analysis

    Question: Let the function f be holomorphic in the open set G. Prove that the function g(z) = \bar{f\left( \bar{z}\right)} is holomorphic in the set   G^\ast =\{\bar{z}: z \in G\}

    Side note: the first bar is over the z and the second bar is over the image of f(z)

    My idea for the proof:
    I want to show that there are first partial derivatives in G* s.t. the partial derivatives are continuous and satisfy the Cauchy Reimann equations
    du \setminus dx=dv\setminus dy  du \setminus dy=-dv \setminus dx

    I differentiate this function on G* w.r.t. to x and y to derive the first partial derivatives.

    (how i do this is f(x+h,y)-f(x,y) over h and f(x,y+h) -f(x,y) over ih)

    However when I do this I get df/dx = du/dx + -i dv/dx
    and df/dy = -dv/dy - i du/dy

    which implies the function is constant.

    Can someone explain to me what I am doing wrong or not seeing?
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  2. #2
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    Quote Originally Posted by arsenicbear View Post
    Question: Let the function f be holomorphic in the open set G. Prove that the function g(z) = \overline{f\left( \overline{z}\right)} is holomorphic in the set   G^\ast =\{\bar{z}: z \in G\}.

    My idea for the proof:
    I want to show that there are first partial derivatives in G* s.t. the partial derivatives are continuous and satisfy the Cauchy Riemann equations
    To do this by using the Cauchy–Riemann equations, let z = x+iy, f(z) = u(x,y) + iv(x,y) and g(z) = U(x,y) + iV(x,y).

    Then g(z) = \overline{f( \overline{z})} = u(x,-y) - iv(x,-y).

    Therefore U(x,y) = u(x,-y) and V(x,y) = -v(x,-y). Now you can find the partial derivatives of U and V, and check that they satisfy the C–R equations.
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  3. #3
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    I think Opalg's solution is correct but requires an advanced theorem; most early in the study of complex analysis have a weaker theorem to work with (easier to prove): if f(z) satisfies the CRE's at z_0 and f_x, f_y (derivatives in the x and y direction) are continuous at z_0 then f is differentiable at z_0. So you probably have to throw in that \overline{f_x{(\overline{z_0})}} and \overline{f_y{(\overline{z_0})}} are continuous as well.

    Alternatively, you can directly apply the definition of the derivative. So for your problem, take any z_0 \in G^*. Then \overline{z_0} \in G \Rightarrow f'(\overline{z_0}) exists. So, \overline{f'(\overline{z_0})} exists (the last step is simply the conjugate of a complex number). Therefore,
    \lim_{z \rightarrow z_0}{\frac{g(z) - g(z_0)}{z - z_0}} <br />
= \lim_{z \rightarrow z_0}{\frac{\overline{f(\overline{z})} - \overline{f(\overline{z_0})}}{z - z_0}}

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