0
$\begingroup$

Let $D_{\rho}$ denote the closed disk of $\mathbb{C}$ , $|z-z_{0}| \leq \rho$ , $\rho >0$ ,$z_{0} \in \mathbb{C}$ are fixed .

Assume that $f$ is analytic on $D_{\rho}$ with taylor series : $$f(z)=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^n$$

Prove that : $$\iint_{D_{\rho}}^{} |f(z)|^2 dx \ dy = \sum_{n=0}^{\infty}|a_{n}|^2 \frac{\rho^{2n +2}}{n+1}$$

  • 0
    What problems did you encounter in the computation? – 2017-01-08
  • 0
    $(z-z_{0})=\rho e^{i\theta}$ , $dx dy = \rho d\rho d \theta$ the problem with $d\theta$. – 2017-01-08
  • 0
    What is the problem with $d\theta$? [As a side note, there's a factor $\pi$ missing on the right hand side, but I don't think that would lead to confusion, only to uncertainty who made the mistake.] – 2017-01-08
  • 0
    I mean the factor $2\pi$. – 2017-01-08
  • 0
    Can you explain your problem more precisely? – 2017-01-08
  • 0
    Is the factor $2\pi$ is missing on the right hand side or not ? – 2017-01-08
  • 0
    It's only a factor $\pi$ that's missing. The $2$ is absorbed in $\dfrac{\rho^{2n+2}}{n+1}$. – 2017-01-08

1 Answers 1

0

Doing the cov $z = z_0 + \rho e^{i\theta}$, $$ \iint_{|z-z_0|\le\rho}|f(z)|^2\,dxdy = \int_0^\rho\!\int_0^{2\pi}|f(z)|^2r\,d\theta dr = \int_0^\rho\!\int_0^{2\pi}\left(\sum_{n=0}^{\infty}a_{n}r^n e^{in\theta}\right) \overline{\left(\sum_{n=0}^{\infty}a_{n}r^n e^{in\theta}\right)}r\,d\theta dr $$ $$ = \int_0^\rho\!\int_0^{2\pi}\left(\sum_{j=0}^{\infty}a_{j}r^j e^{ij\theta}\right) \left(\sum_{k=0}^{\infty}\overline{a_{k}}r^k e^{-ik\theta}\right)r\,d\theta dr = $$ $$ = \int_0^\rho\!\int_0^{2\pi}\left(\sum_{n=0}^{\infty} \left(\sum_{j+k=n}a_{j}\overline{a_{k}}e^{i(j-k)\theta}\right)r^{n+1}\right)d\theta dr. $$ Because of the periodicity of imaginary exponential, the terms with $j\ne k$ give $0$. On the other hand, $j = k$ implies $e^{i(j-k)\theta} = 1$ and $j + k$ even, so: $$ \iint_{|z-z_0|\le\rho}|f(z)|^2\,dxdy = 2\pi\int_0^{\rho}\sum_{n=0}^{\infty} |a_n|^2 r^{2n+1}dr = 2\pi\sum_{n=0}^{\infty} |a_n|^2\frac{\rho^{2n+2}}{2n+2}. $$