# Entire function with special conditions [closed]

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

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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

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