# Fourier series via complex analysis

1. Apr 8, 2008

### Mystic998

1. The problem statement, all variables and given/known data

Show that f is 2-pi periodic and analytic on the strip $\vert Im(z) \vert < \eta$, iff it has a Fourier expansion $f(z) = \sum_{n = -\infty}^{\infty} a_{n}z^{n}$, and that $a_n = \frac{1}{2 \pi i} \int_{0}^{2\pi} e^{-inx}f(x) dx$. Also, there's something about the lim sup of the modulus nth roots of the coefficients being less than $e^{-\eta}$. But I think that follows pretty obviously from the Laurent series existing.

2. Relevant equations

3. The attempt at a solution

The direction assuming it has a Fourier expansion is pretty easy. Just set $\omega = e^{iz}$. Then the series with $\omega$ is analytic on the strip, and that it's 2-pi periodic is clear.

For the other direction I think the idea here is to find a Laurent series for f(z) (or at least that agrees with f on a "big enough" subset of its domain) that's actually analytic on the annulus $e^{-\eta} < \vert z \vert < e^{\eta}$. Then you just use that Laurent series with the variable replaced by $e^{iz}$.

However there doesn't seem to be any way to get such a series that I can see. I've tried a lot of ideas: I've tried looking for a Laurent series for f in the intersection of the annulus with the strip, I've tried finding an appropriate Laurent series for f for an annulus contained in the strip, and I can't get those to work. Also I've tried cheating and just appealing to the Fourier series existing for a real valued function and then using the fact that if analytic functions agree on a "big enough" part of their domain, they have to be the same, but that just seems like a cop out and not very convincing.

So, I'd appreciate any hints or help. Thanks.

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