Prove that $\frac{1}{2\pi}\int_0^{2\pi}|p(e^{i\theta})|^2\,d\theta=\sum_{n=0}^N|a_n|^2.$

Question

Let , $\displaystyle p(z)=\sum_{n=0}^Na_nz^n$ be a complex polynomial. Then show that , $$\frac{1}{2\pi}\int_0^{2\pi}|p(e^{i\theta})|^2\,d\theta=\sum_{n=0}^N|a_n|^2.$$

$$\frac{1}{2\pi}\int_0^{2\pi}|p(e^{i\theta})|^2\,d\theta=\frac{1}{2\pi}\int_0^{2\pi}|\sum_0^Na_ne^{in\theta}|\,d\theta\le \left(\sum_0^N|a_n|\right)^2$$But from here how I can prove the result ?

Answer 1

Hint:

Write $$\left|p(e^{i\theta})\right|^{2}=\left|\sum_{n=0}^{N}a_{n}e^{in\theta}\right|^{2}=\left(\sum_{n=0}^{N}a_{n}e^{in\theta}\right)\overline{\left(\sum_{m=0}^{N}a_{m}e^{im\theta}\right)}.$$ Expanding the above product we obtain the double sum $$\left|p(e^{i\theta})\right|^{2}=\sum_{n,m=0}^{N}a_{n}\overline{a_{m}}e^{i(n-m)\theta}.$$ Now, what happens to each term individually when we integrate?