$\sum_{n=0}^N |b_n|^2 r^{2n}=\int_{-\pi}^{\pi}|f(z_0+re^{in\theta})|^2d\theta$

Question

This problem comes from page 110 of rudin's real and complex analysis:

The problem I am having is with the equality

$$\sum_{n=0}^N |b_n|^2r^{2n}=\int_{-\pi}^{\pi}|f(z_0+re^{in\theta})|^2d\theta$$

Working the LHS you get some work that looks like this:

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(z_0+re^{in\theta})|^2d\theta = \frac{1}{2\pi}\int_{-\pi}^{\pi}|\sum\limits_{n=0}^{N}b_n r^ne^{in\theta}|^2d\theta$$

If by some gift of god you could show that

$$|\sum\limits_{n=0}^{N}b_n r^ne^{in\theta}|^2=\sum_{n=0}^N |b_n r^ne^{in\theta}|^2$$ then the equality follows obviously. Any ideas guys?

Answer 1

Something less than the intervention of a deity:

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}|\sum_{n=0}^{N}b_n r^ne^{in\theta}|^2d\theta = \frac{1}{2\pi}\int_{-\pi}^{\pi}(\sum_{m=0}^{N}b_m r^me^{im\theta})\overline {(\sum_{n=0}^{N}b_n r^ne^{in\theta})}\, d\theta$$ $$=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left (\sum_{m=0}^{N}\sum_{n=0}^{N} b_m\overline {b_n}r^{m+n} e^{i(m-n)\theta}\right)\,d\theta $$