$f(z)=\sum_{n=0}^\infty a_n(z-a)^n$ converges for $|z-a|

Question

Suppose $f(z)=\sum_{n=0}^\infty a_n(z-a)^n$ converges for $|z-a|<s$ .

Then how to show that for $0<r<s$ , $\dfrac 1{2\pi}\int_0^{2\pi} |f(a+re^{it})|^2 dt=\sum_{n=0}^\infty |a_n|^2r^{2n}$ ?

Answer 1

This equality is called Parseval's identity.

As $f(z)=\sum_{n=0}^\infty a_n(z-a)^n$ for $|z-a|<s$, if we take $r<s$, then $|a+re^{i\theta}-a|=r\leq s$, so:

$$\int_0^{2\pi} |f(a+re^{i\theta}|^2\ d\theta=\int_0^{2\pi} f(a+re^{i\theta})\cdot \overline{f(a+re^{i\theta})}\ d\theta=\int_0^{2\pi} \left(\sum_{n=0}^\infty a_nr^ne^{in\theta}\right)\cdot \left(\sum_{n=0}^\infty \overline{a_n}r^ne^{-in\theta}\right)\ d\theta=$$ Now, using the Cauchy product, we have $$ = \int_0^{2\pi}\sum_{n,m=0}^\infty a_n\cdot \overline{a_m}\cdot r^{n+m}e^{i(n-m)\theta}\ d\theta=\sum_{n,m=0}^\infty a_n\cdot \overline{a_m}\cdot r^{n+m}\int_0^{2\pi} e^{i(n-m)\theta}\ d\theta$$

As $$\int_0^{2\pi} e^{i(n-m)\theta}\ d\theta=\left\{\begin{array}{c} 2\pi, \ n=m \\ 0, \ n\neq m \end{array} \right. $$ you get

$$\int_0^{2\pi} |f(a+re^{i\theta}|^2\ d\theta = 2\pi \sum_{n=0}^\infty |a_n|^2 r^{2n}$$

Answer 2

Observe \begin{align} |f(a+re^{it})|^2 =\left|\sum^\infty_{n=0} a_nr^ne^{int}\right|^2 = \sum^\infty_{n=0}\sum^\infty_{m=0}a_n\bar a_m r^{n+m}e^{i(n-m)t} \end{align} for all $r<s$ which means \begin{align} \frac{1}{2\pi}\int^{2\pi}_0\sum^\infty_{n=0}\sum^\infty_{m=0}a_n\bar a_m r^{n+m}e^{i(n-m)t} \ dt =&\ \sum^\infty_{n=0}\sum^\infty_{m=0}a_n\bar a_m r^{n+m}\frac{1}{2\pi}\int^{2\pi}_0e^{i(n-m)t} \ dt \\ =&\ \sum^\infty_{n=0}\sum^\infty_{m=0}a_n\bar a_m r^{n+m} \delta_{n, m} \\ =&\ \sum^\infty_{n=0}|a_n|^2r^{2n}. \end{align}