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Math Help - Complex Analysis

  1. #1
    Senior Member Dinkydoe's Avatar
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    Complex Analysis

    Let f be a continuous function on D=\left\{z:|z|=1\right\}. Laat \gamma:[0,2\pi]\to D, defined by \gamma(t)=e^{it}.

    Define the function F as follows:
    F(z) = f(z) if |z|=1
    F(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta if |z|< 1

    Is F continuous on \overline{D}(0,1) [Hint:Consider f(z)=\overline{z}]

    How would one go to tackle this problem?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    Let f be a continuous function on D=\left\{z:|z|=1\right\}. Laat \gamma:[0,2\pi]\to D, defined by \gamma(t)=e^{it}.

    Define the function F as follows:
    F(z) = f(z) if |z|=1
    F(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta if |z|< 1

    Is F continuous on \overline{D}(0,1) [Hint:Consider f(z)=\overline{z}]

    How would one go to tackle this problem?
    Do what the hint says, take f(z) = \overline{z}. If |z| = 1 then F(z) = \overline{z}.

    If |z| < 1 then F(z) = \frac{1}{2\pi i}\int_\gamma \frac{\overline{\zeta}}{\zeta-z}d\zeta. But when |\zeta|=1, \overline{\zeta} = \zeta^{-1}. So we can write F(z) as \frac{1}{2\pi i}\int_\gamma \frac1{\zeta(\zeta-z)}d\zeta = -\frac1z+\frac1z=0 (using the residue theorem to evaluate the integral, since the integrand is now analytic except for poles at \zeta=0 and \zeta=z).

    That shows that F(z) is not continuous anywhere on \gamma.
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