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Let $f$ be holomorphic on an open set containing the closed unit disc $\bar{D}$, $D=\{z\in\Bbb C:|z|<1\}$. Let $\gamma$ be the unit circle, parametrized counterclockwise. Prove that, for any $a\in\Bbb C$, $|a|\neq1$, we have $$\frac{1}{2\pi i}\int_\gamma\frac{\overline{f(z)}}{(z-a)}dz=\begin{cases}\overline{f(0)} &|a|<1\\[6pt] \overline{f(0)}-\overline{f(1/\bar{a})}&|a|>1.\end{cases}$$

I am completely stuck on this problem. Somehow, I want to use Cauchy's formula, but I know the conjugate of $f$ is not necessarily holomorphic.

Any help?

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  • $\begingroup$ The intersection of holomorphic and anti-holomorphic functions consists of constants... $\endgroup$ Dec 23, 2014 at 21:36
  • $\begingroup$ @paulgarrett I dont get it. Could you say a little more about it? $\endgroup$
    – YYF
    Dec 23, 2014 at 21:45
  • $\begingroup$ Let $f(z)=u(z)+iv(z)$ with $u$ and $v$ real valued functions. By $\overline{f(z)}$, do you mean $u(z)-iv(z)$ or $u(\overline z)-iv(\overline z)$. $\endgroup$ Dec 23, 2014 at 21:58
  • $\begingroup$ The general idea is that the Cauchy formula "projects to holomorphic functions". On a disk, an anti-holomorphic function has a power series expansion in $\overline{(z-z_o)}$, and all but the constant term project to $0$. $\endgroup$ Dec 23, 2014 at 22:01
  • $\begingroup$ @TimRaczkowski I think it should be the former one. $\endgroup$
    – YYF
    Dec 23, 2014 at 22:17

1 Answer 1

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Hint. If $\,f(z)=\sum_{n=0}^\infty a_nz^n,$ then $$ \overline{f(z)}=\sum_{n=0}^\infty \overline{a_n}\,\overline{z}^n, $$ and if $\lvert z\rvert=1$, then $\overline{z}=z^{-1}$, and thus $$ \overline{f(z)}=\sum_{n=0}^\infty \overline{a_n}\,{z}^{-n}. $$ Hence $$ \frac{1}{2\pi i}\int_\gamma\frac{\overline{f(z)}\,dz}{z-a}=\frac{1}{2\pi i}\int_\gamma \sum_{n=0}^\infty \frac{\overline{a_n}\,dz}{z^{n}(z-a)}=\frac{1}{2\pi i} \sum_{n=0}^\infty\int_\gamma \frac{\overline{a_n}\,dz}{z^{n}(z-a)}. $$

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