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Hi all, here's my question which I have no idea how to approach. Fix a complex number q such that |q| < 1. Describe all entire functions f such that f(z)/f(qz) is a linear function of z.

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closed as too localized by Igor Rivin, Andrés E. Caicedo, S. Carnahan Jun 28 '11 at 7:16

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Look at the power series at the origin... – Igor Rivin Jun 27 '11 at 15:29
Please see the FAQ for a list of sites where your question may be more appropriate. – S. Carnahan Jun 28 '11 at 7:17
On the one hand, it does look like a homework question. On the other, it is probably a graduate level question, albeit not research level, which the FAQ says is fine. – Richard Rast Jun 29 '11 at 1:04
up vote 1 down vote accepted

Consider zeros of $f$. If $f(0)=0$, we can, for some n, write $f(z)=z^n g(z)$, $f(qz)=z^n q^n g(qz)$, and we find that $g(z)/g(qz)$ is also a linear function. Now let us say $g(z)/g(qz)=az+b$. By plugging in $z=0$, we find $b=1$. Moreover, $g(-1/a)=0$. Since $az+1$ has no poles, we recursively find $g(-1/(qa))=g(-1/(q^2a))=...=0$. Now let $h(z)=\prod_{n=0}^\infty (aq^nz+1)$. Since $|q|<1$, this is a convergent product. Define $u(z)=g(z)/h(z)$. We find that $u(z)/u(qz)=1$. This yields $u(1)=u(1/q)=u(1/q^2)$=...=$u(0)$, which implies that u is constant.

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at the point $z=−b/(qa)$, $g(z)/g(qz)$ has a pole ... or $g(z)=0$. – Gerald Edgar Jun 27 '11 at 20:33
Yes. I think I fixed the argument. – Michael Renardy Jun 27 '11 at 21:29

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