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

Question

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 $\int \int_{\mathbb{D}} \frac{\partial f}{\partial \bar{z}}\frac{1}{z-\zeta}dx dy$

Answer 1

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.

Answer 2

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.