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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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    $\begingroup$ Look at the power series at the origin... $\endgroup$
    – Igor Rivin
    Jun 27, 2011 at 15:29
  • $\begingroup$ Please see the FAQ for a list of sites where your question may be more appropriate. $\endgroup$
    – S. Carnahan
    Jun 28, 2011 at 7:17
  • $\begingroup$ 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. $\endgroup$ Jun 29, 2011 at 1:04

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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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  • $\begingroup$ at the point $z=−b/(qa)$, $g(z)/g(qz)$ has a pole ... or $g(z)=0$. $\endgroup$ Jun 27, 2011 at 20:33
  • $\begingroup$ Yes. I think I fixed the argument. $\endgroup$ Jun 27, 2011 at 21:29

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