Cambodia just cremated ex-King Norodom Sihanouk with great fanfare. But in Cambodia death and tourism are never far apart. Even the Killing Fields have been "sold" to the Japanese so they can package it better for tourists.



via Firstpost http://www.firstpost.com/world/death-tourism-in-phnom-penh-of-dead-kings-and-killing-fields-614512.html

Firstpost johalputt Roy

04:10

For which $a \in \mathbb{C} -\ \{0,1\}$ does the equation $f(z) = f(az) $ has a non constant solution $f$ with $f$ being analytical in a neighborhood of $0$.

My attempt:

First, we can see that any such solution must satisfy:

$f(z)=f(a^kz)$ for all $k \in \mathbb{N} $.

If $|a|<1$:

The series $z_{k} = a^k$ converges to 0 which is an accumulation point, and $f(z_i)=f(z_j)$ for all $i, j\in \mathbb{N} $. Thus $f$ must be constant.

If $|a|=1$:

For all $a \neq 1$ , $f$ must be constant on any circle around $0$, so again $f$ must be constant.

My quesions are:

Am I correct with my consclusions?

Also, I'm stuck in the case where $|a|>1$. Any ideas?

Thanks

via Recent Questions - Mathematics - Stack Exchange http://math.stackexchange.com/questions/291287/for-which-a-does-the-equation-fz-faz-has-a-non-constant-solution-f

johalputt Recent Questions - Mathematics - Stack Exchange Roy

03:47