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Consider a function of a complex variable $f(z):\mathbb{C}\to\mathbb{C}$ with a periodicity condition

$$f(z+a)=f(z)~~~,~~~a\in\mathbb{R},$$

What would be a convenient basis of functions to expand $f(z)$ on that will take its periodicity into account?

The Fourier basis functions naturally accommodate periodicity and can be applied for complex valued functions of a real variable, but what if $f(z)$ is a function of complex variable? Do convenient basis functions exist for that case? Thanks for any suggestion.

Kagaratsch
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  • The answer seems to be here: https://math.stackexchange.com/questions/2580735/fourier-series-expanding-of-holomorphic-functions – Kagaratsch Jan 30 '20 at 22:30

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The Fourier series coefficients still work, though they don't preserve the Hermitic Symmetry. To prove this, let $$f(z)=\sum_{n=-\infty}^\infty a_ne^{{2\pi n\over a}z}$$therefore $$a_n={1\over a}\int_0^af(z)e^{-{2\pi n\over a}z}dz$$where for $f(z):\Bbb R\to\Bbb R$ we have $$a_{-n}=a^*_n$$

Mostafa Ayaz
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  • Thank you for your answer. May I ask, since the equation for $a_n$ you mention only involves a real valued slice from the domain $0\leq$Re$(z)<a$, $-\infty<$Im$(z)<\infty$, how can we be sure that the expansion properly works away from the real axis? – Kagaratsch Jan 30 '20 at 02:58
  • Good point. Actually it works if the function $f(z)$ is analytic over the real line as we had such cases for real-valued functions $f$. Except that, I don't see any basis as the function may not have a periodic representation at all. – Mostafa Ayaz Jan 30 '20 at 03:01
  • So, you are saying, this basis is not appropriate for $f(z)$ as defined in the question after all? – Kagaratsch Jan 30 '20 at 03:02
  • In its most general form, yes. For example let $$f(z)=\begin{cases}z&,\quad 0\le \Re(z)<1,\Re(z)\in\Bbb Q\0&,\quad 0\le \Re(z)<1,\Re(z)\notin\Bbb Q\end{cases}$$where $a=1$. – Mostafa Ayaz Jan 30 '20 at 03:07
  • Ok, but this is a rather peculiar example. What if $f(z)$ is infinitely smoothly differentiable? I tend to think, in that case, since the Fourier expansion clearly holds along the real slice, its analytic continuation into the complex direction is smooth and must be given by the same function. Would you agree with that? – Kagaratsch Jan 30 '20 at 05:03
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    Yes but when walking through the complex domain, complex coefficients begin to appear in the Fourier expansion. For example a function $$f(x)=k\sin x$$always has a Fourier transform.whether or not $k\in \Bbb R$. – Mostafa Ayaz Jan 30 '20 at 14:08