# Fourier coefficients

## Homework Statement

Let ##f## be a ##2\pi## periodic function. Let ##\hat{f}(n)## be the fourier coefficient of ##f## defined by
$$\hat{f}(n)=\frac{1}{2\pi}\int_{a}^{b}f(x)e^{-inx}dx.$$
for ##n\in\mathbb{N}##. If ##\overline{\hat{f}(n)}=\hat{f}(-n)## show that ##f## is real valued.

## The Attempt at a Solution

Suppose for contradiction that ##f## is not real valued. Then
$$\overline{\hat{f}(n)}=\frac{1}{2\pi}\overline{\int_{a}^{b}f(x)\cos nxdx}=\frac{1}{2\pi}\left[\overline{\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx}\right]\\ =\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx-i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]$$
and
$$\hat{f}(-n)=\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]$$
so ##\hat{f}(n)\not=\hat{f}(-n)## a contradiction. Can anyone verify my proof for me? Thanks!

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Note that $f$ is real valued by assumption!