# Can Fourier series be applied to a function of a complex variable?

If $$f(x)=\displaystyle\sum_{n=-\infty}^\infty c_{n,1}e^{inx}$$ (complex Fourier series) and $$c_{n,1}=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^\pi f(x)e^{-inx} \, dx$$ where $$x\in [-\pi ,\pi]$$, is it true that $$f(zx)=\displaystyle\sum_{n=-\infty}^\infty c_{n,2}e^{inzx}$$ and $$c_{n,2}=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^\pi f(zx)e^{-inzx} \, dx$$ where $$x\in [-\pi ,\pi]$$ and $$z\in \mathbb C$$?

• If the values of $f(zx)$ are not determined by the values of $f(x)$, would you expect to be able to determine the values of $f(zx)$ from the Fourier series of $f(x)$? – DisintegratingByParts May 30 at 14:19

That's not even true for real $$z$$, much less complex $$z$$. Try it with $$z=2$$ and you'll immediately see why. That said, I think these might shed light on the question you're trying to get at.

If $$f(zx)$$ is integrable on $$[-\pi,\pi]$$, this statement is certainly true: $$f(zx) = \sum_{n=-\infty}^\infty c_{n,2}e^{inx}\;\;\;\;;\;\;\;\; c_{n,2} = \frac{1}{2\pi} \int_{-\pi}^\pi f(zx)e^{inx}dx,$$ but that's because $$f(zx)$$ is just another integrable complex-valued function of a real variable and Fourier series work on all such functions.

On the other hand, this almost never works: $$f(z) = \sum_{n = -\infty}^\infty c_{n,3}e^{inz}\;\;\;\;;\;\;\;\; c_{n,3} = \frac{1}{2\pi}\int_{-\pi}^\pi f(z)e^{inz} dz$$ for the simple reason that if $$\mathrm{Im}[z]\ne0$$, the sum has a term that scales as $$\exp(-n\mathrm{Im}[z])$$, so in general it will diverge as $$n\rightarrow \pm \infty$$. For example, if we use the periodic function $$f(z) = \pi^2-(z\mod 2\pi)^2$$, which is defined for all $$z$$, we get $$c_{n,3} = 2(-1)^n/n^2$$ for $$n\ne 0$$ and $$c_{0,3} = 2\pi^2/3$$. If we try the Fourier series sum and let $$z = x+iy$$, we get $$\begin{multline} f(z) = \sum_{n = -\infty}^\infty c_{n,3}e^{inz} = \frac{2\pi^2}{3} +4\sum_{n = 1}^\infty \frac{(-1)^n}{n^2}\cos(nz) \\ = \frac{2\pi^2}{3} +4\sum_{n = 1}^\infty \frac{(-1)^n\cosh(ny)}{n^2}\left[\cos(nx) + i\sin(nx)\tanh(ny)\right], \end{multline}$$ which clearly diverges if $$y\ne 0$$, since $$\cosh(ny)/n^2$$ is unbounded. For this to work with $$y \ne 0$$, $$f(z)$$ must not only have period $$2\pi$$, but also be analytic on $$\mathbb R$$, a much, much stronger condition than simple periodicity.

• That's a really helpful answer. In the wrong procedure, I just multiplied every $x$ by $z$ (apart from the differential). Now I see it didn't make sense. – Poder Rac May 30 at 15:16