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Let $f(x)=c_0+c_1e^{i\theta}+c_2e^{2i\theta}+...+c_ne^{ni\theta}$ where $c_k\in \mathbb R$. We need to show $$\int_{0}^{2\pi}f(e^{i\theta})\overline {f(e^{i\theta})} d\theta =2\pi\sum_{k=0}^{n} c_k$$ So far I have tried expanding this to: $$\alpha=[c_0+c_1cos(\theta)+c_2cos(2\theta)...+c_ncos(n\theta)], \beta=[c_1sin(\theta)+c_2sin(2\theta)...+c_nsin(n\theta)]$$ Thus, $f\bar f = (\alpha+i\beta)(\alpha-i\beta)=\alpha^2+\beta^2$

This isn't much progress, but whenever I try to simplify this expression I fail. Could someone at least point me in the right direction?

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up vote 1 down vote accepted

The basic idea behind this is to first observe that $$\overline{f(e^{i\theta})} = f(\overline{e^{i\theta}}) = f(e^{-i\theta}).$$ Then we note that the product is $$f(e^{i\theta}) \overline{f(e^{i\theta})} = \sum_{k=0}^n c_k e^{k i\theta} \sum_{m=0}^n c_m e^{-m i \theta} = \sum_{k=0}^n \sum_{m=0}^n c_k c_m e^{(k-m)i\theta}.$$ Then consider integrating a representative term of this sum over $\theta \in [0,2\pi]$: it vanishes if and only if $k \ne m$. Therefore, the integral consists only of terms for which $k = m$.

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