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Let $f$ be holomorphic in a neighborhood of the closure of the unit disk $D$.

Show that $\forall |z|\le 1$

$f(z)(1-|z|^2)= \frac{1}{2\pi i}\int _{|u|=1} \frac{1-\overline{z}u}{u-z}f(u)du$.

This is what I tried to do, but I'm a little confused. Well.. Since the curve $|u|=1$ is contained in the domain of the analytic function , we can use Cauchy integral formula. Using Cauchy integral formula, and then multiplying by $(1-|z|^2)$ to both sides, we get the following equality:$$ f\left( z \right)\left( {1 - |z|^2 } \right) = \frac{1} {{2\pi i}}\int\limits_{|u| = 1} {\frac{{f\left( u \right)}} {{u - z}}\left( {1 - |z|^2 } \right)du} $$ So it's remain to show that: $$ \int\limits_{|u| = 1} {\frac{{f\left( u \right)}} {{u - z}}\left( {1 - |z|^2 } \right)du} = \int\limits_{|u| = 1} {\frac{{f\left( u \right)}} {{u - z}}\left( {1 - \overline z u} \right)du} $$ Or equivalently: $$ 0 = \int\limits_{|u| = 1} {\frac{{f\left( u \right)}} {{u - z}}\left( {\overline z u - |z|^2 } \right)du} = \int\limits_{|u| = 1} {\frac{{f\left( u \right)}} {{u - z}}\overline z \left( {u - z} \right)du = } \int\limits_{|u| = 1} {f\left( u \right)\overline z du = } \overline z \int\limits_{|u| = 1} {f\left( u \right)du} $$

But something has to be wrong here

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    $\begingroup$ $f$ is homomorphic, so your final integral is zero by Cauchy's Theorem! $\endgroup$ – user123123 Oct 29 '12 at 17:20
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    $\begingroup$ Perhaps I'm missing something but me thinks you already have it: $$\oint_{|u|=1}f(u)\,du=0$$ since $\,f\,$ is analytic on and within the unit circle! $\endgroup$ – DonAntonio Oct 29 '12 at 17:20
  • $\begingroup$ Oh you are completely right! Thanks $\endgroup$ – Daniel Oct 29 '12 at 17:22

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