# Double Integral Formula for Holomorphic Function on the Unit Disc (Complex Plain)

While studying some complex analysis , I encountered the following problem:

"Let $f$ be a holomorphic function on the open unit disc $\mathbb{D}$.Prove that for every $\zeta \in \mathbb{D}$ the following formula is valid: $f(\zeta)=\frac{1}{\pi} \int \int_{\mathbb{D}}\frac{f(z)}{(1-\bar{z}\zeta)^2}dxdy$ "

I tried this by using Green's Theorem and a Cauchy-Green theorem, but i ended up with a double integral expression of the form $\int \int_{\mathbb{D}} \frac{\partial f}{\partial \bar{z}}\frac{1}{z-\zeta}dx dy$

• What do you get with Green's theorem exactly ? – reuns Nov 14 '16 at 0:18
• I consider the Cauchy Integral Formula as a contour integral and i use Green's Theorem to get the statement that i posted under the question. It is a bit long to post it in this comment. I used Sarason's book to find a similar statement but nothing occured . – Beslikas Thanos Nov 14 '16 at 0:22
• With Green's theorem you are supposed to write $\frac{f(z)}{(1-\overline{z}\zeta)^2}$ as $\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}$ and obtain a contour integral – reuns Nov 14 '16 at 0:25
• I did this but on $\oint \frac{f(z)}{z-\zeta}dz$ so i guess that's why i was wrong. I will try your hint, thank you! – Beslikas Thanos Nov 14 '16 at 0:28
• What do you mean with $\oint \frac{f(z)}{z-\zeta} dz$ ? And do you know Cauchy's integral formula ? – reuns Nov 14 '16 at 0:30

Sketch: Assume first $f$ is holomorphic on a neighborhood of $\overline {\mathbb D}.$ Write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Recall that

$$\frac{1}{(1-u)^2} =\sum_{n=0}^{\infty}(n+1)u^n.$$

Hence

$$\frac{1}{(1-\bar z{\zeta})^2} =\sum_{n=0}^{\infty}(n+1)(\bar z{\zeta})^n.$$

Now compute

$$\frac{1}{\pi}\int_{\mathbb D}\left (\sum_{n=0}^{\infty}a_nz^n \right )\left (\sum_{n=0}^{\infty}(n+1)(\bar z{\zeta})^n \right )\, dA(z)$$

using polar coordinates and orthogonality of the functions $z^n$ with respect to area measure on $\mathbb D.$ This works out nicely.

• This should be tractable. +1 – Mark Viola Nov 14 '16 at 1:45
• Never thought of taking the series representation! – Beslikas Thanos Nov 14 '16 at 8:00

This can be solved with Green's Formula also. Using the Cauchy Integral Formula one has $$f(\zeta)=\frac{1}{2\pi i}\int_{|z|=1}{\frac{f(z)}{z-\zeta}}dz=\int_{|z|=1} \frac{f(z)\bar{z}}{1-\bar{z}\zeta}dz$$ The last statement can be justified by the fact that $z=\frac{1}{\bar{z}}$ when $|z|=1$ . Now one can use Green's theorem . After some fun one has $$f(\zeta)=\frac{1}{\pi}\int \int_{\mathbb{D}}\frac{\partial{\frac{f(z)\bar{z}}{1-\bar{z}\zeta}}}{\partial{\bar{z}}}dxdy=\frac{1}{\pi}\int \int_{\mathbb{D}}\frac{f(z)}{(1-\bar{z}\zeta)^2 }dxdy.$$ My effort was still good, i just did not consider the simple step $$z=\frac{1}{\bar{z}}$$ when $|z|=1$. Thanks also for the answer using the series representation it was a simple and beautiful way to solve this excercise.