Suppose a complex function $f(z)$ depends only on $z$, and satisfies the quasi-periodicity in both directions: $$f(z+ a_x)= e^{i \theta_{a_x}} f(z)$$ $$f(z+ i a_y)= e^{i \theta_{a_y}} f(z)$$ where $\theta_{a_x}$ and $\theta_{a_y}$ are real functions of $a_x$ and $a_y$ respectively. One can expand $f(z)$ using Laurent expansion $\sum_n c_n z^n$, but $z^n$ is not quasi-periodic. My question is: does there exist a basis which respects the quasi-periodicity and can expand arbitrary quasi-periodic complex function?
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$\begingroup$ I don't think the word quasi-periodic is adapted. The only meromorphic functions such that $f(z+a)=A f(z), f(z+b)=Bf(z)$ for some $a/b\not \in \Bbb{R}$ are the quotients of theta functions ($\theta(z)=\sum_n e^{2i\pi n u z} e^{-\pi n^2 a/b}$ satisfies the same thus $f(z)/\theta(z)$ is doubly periodic, then construct a quotient of theta functions having the same zeros/poles) $\endgroup$– reunsCommented Feb 6, 2020 at 20:01
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Bloch's theorem says that $f(z)=e^{iθz/a}u(z)$, with $u(z+a)=u(z)$. Then $u(z)=\sum_{n=-\infty}^\infty c_n \xi^n$ has a series expansion in powers of $\xi=e^{2\pi iz/a}$, which can be seen as a Fourier or Laurent series (see Laurent series yields Fourier series).
answered Feb 5, 2020 at 21:55
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$\begingroup$ thank you! I agree with your first part of the answer, using the Bloch theorem. However, I found I formulated the question too sloppy. What I mean is that $f(z)$ is quasi-periodic in both directions on torus. So $\xi= e^{2\pi i z/a}$ is not the transformation I really want (which is periodic only in one direction). I re-edited the question. $\endgroup$ Commented Feb 6, 2020 at 14:08
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$\begingroup$ For $z=x+iy$ you have a function $f(x,y)$ on $\mathbb{R}^2$, which satisfies $f(x,y)=e^{i\theta_x x/a_x}e^{i\theta_y y/_y}u(x,y)$ and $u(x,y)=\sum_{m,n}c_{mn}\xi_x^m\xi_y^n$, with $\xi_x=e^{2\pi i x/a_x}$, $\xi_y=e^{2\pi i y/a_y}$. $\endgroup$ Commented Feb 6, 2020 at 15:15