Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

closed as too localized by Igor Rivin, Andres 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.

1 Answer

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.

share|improve this answer
    
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
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.