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.
3
-
2$\begingroup$ Look at the power series at the origin... $\endgroup$– Igor RivinCommented 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 ♦Commented 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. $\endgroup$– Richard RastCommented Jun 29, 2011 at 1:04
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Jun 27, 2011 at 19:27
-
$\begingroup$ at the point $z=−b/(qa)$, $g(z)/g(qz)$ has a pole ... or $g(z)=0$. $\endgroup$ Commented Jun 27, 2011 at 20:33
-
$\begingroup$ Yes. I think I fixed the argument. $\endgroup$ Commented Jun 27, 2011 at 21:29