$\quad$ Using Cauchy's integral theorem, write down the value of a holomorphic function $f(z)$ where $|z|\lt1$ in terms of a contour integral around the unit circle, $\zeta=e^{i\theta}$.
$\quad$ By considering the point $1/\overline z$, or otherwise, show that $$f(z)=\dfrac1{2\pi}\int_0^{2\pi}f(\zeta)\dfrac{1-|z|^2}{|\zeta-z|^2}\,\mathrm d\theta.$$

Any help with the above question would be most appreciated!

It is the first part of a (so-called) long / Section II question - given this part, I can (and have) completed the rest of the question.

It is very fiddly and I just can't quite get the final answer out correctly.

(I have written out Cauchy's integral formula for the first part shown, just the final manipulation is what I need help with.)

Thanks very much for your time! :)

PS - if this is a duplicate, then I'm sorry, but this can't really be given a unique title, and I'm not going to check every single question on Cauchy's integral formula! :P


Start with the usual integral formula, and expand $d\zeta$ using $\zeta = e^{i\vartheta}$:

$$f(z) = \frac{1}{2\pi i}\int_{\lvert \zeta\rvert = 1} \frac{f(\zeta)}{\zeta-z}\,d\zeta = \frac{1}{2\pi i}\int_0^{2\pi} \frac{f(\zeta)}{\zeta-z}i\zeta\,d\vartheta = \frac{1}{2\pi}\int_0^{2\pi} \frac{f(\zeta)}{(\zeta-z)\overline{\zeta}}\,d\vartheta.\tag{1}$$

Now suppose $f$ is holomorphic in a neighbourhood of the closed unit disk (the general case follows by taking a limit), fix $w$ with $\lvert w\rvert < 1$, and apply $(1)$ to the function

$$z \mapsto \frac{f(z)}{1 - z\overline{w}}.$$

You obtain

$$\begin{align} \frac{f(z)}{1-z\overline{w}} &= \frac{1}{2\pi}\int_0^{2\pi} \frac{f(\zeta)}{(1-\zeta\overline{w})\overline{\zeta}(\zeta-z)}\,d\vartheta\\ &= \frac{1}{2\pi} \int_0^{2\pi} \frac{f(\zeta)}{(\overline{\zeta}-\overline{w})(\zeta-z)}\,d\vartheta. \end{align}$$

Choosing $z = w$ gives

$$\frac{f(w)}{1-\lvert w\rvert^2} = \frac{1}{2\pi}\int_0^{2\pi} \frac{f(\zeta)}{\lvert \zeta-w\rvert^2}\,d\vartheta,$$

which after multiplication with $1-\lvert w\rvert^2$ and renaming $w$ to $z$ becomes the desired

$$f(z) = \frac{1}{2\pi}\int_0^{2\pi} f(\zeta)\frac{1-\lvert z\rvert^2}{\lvert\zeta-z\rvert^2}\,d\vartheta.$$

  • $\begingroup$ Sorry, I don't know what you mean by applying $(1)$ to the function $z \mapsto \frac{f(z)}{1 - z\overline{w}}$. I also forgot to add that there's an extra hint, saying to recall that $\zeta = 1/ \overline{\zeta}$ for $\zeta$ on the unit circle, and hence $${\zeta \over {\zeta - 1/ \overline{z}}} = -{\overline{z} \over {\overline{\zeta} - \overline{z}}}.$$ I can't see how to get the bit with the mod squared though... $\endgroup$ – Sam T May 2 '14 at 21:52
  • $\begingroup$ Formula $(1)$ holds for all $f$ holomorphic in a neighbourhood of the closed unit disk (or, by taking limits, holomorphic on the unit disk and continuous on the closure). Now for your given $f$, consider $g(z)=\frac{f(z)}{1-z\overline{w}}$. Apply $(1)$ to that, so $$g(z)=\frac{1}{2\pi}\int_0^{2\pi}\frac{g(\zeta)}{(\zeta-z)\overline{\zeta}}\,d\vartheta.$$ Insert what $g$ is, $$\frac{f(z)}{1-z\overline{w}}=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(\zeta)}{(\zeta-z)\overline{\zeta}(1-\zeta\overline{w})}\,d\vartheta.$$ Use $\lvert\zeta\rvert=1$ and choose an appropriate $z$ to obtain the desired formula. $\endgroup$ – Daniel Fischer May 2 '14 at 22:00
  • $\begingroup$ Do you mean "choose an appropriate $w$", ie $w = z$, and then multiply through by $1 - z \overline z = 1 - |z|^2$ - this gives the required result as $\overline{\zeta} \zeta = 1$ (in the denominator)? $\endgroup$ – Sam T May 2 '14 at 22:07
  • $\begingroup$ Well, we chose $w$ first, so we pick $z = w$ and then rename. But yes, that's it. $\endgroup$ – Daniel Fischer May 2 '14 at 22:09
  • $\begingroup$ Ok yeah, I see. Thank you very much! :) $\endgroup$ – Sam T May 2 '14 at 22:23

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