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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 $z\in \mathbb C$?

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    $\begingroup$ 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 $f(x)$? $\endgroup$ – DisintegratingByParts May 30 at 14:19
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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.

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  • $\begingroup$ 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. $\endgroup$ – Poder Rac May 30 at 15:16

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