Find all analytic functions satisfying the following condition

Question

Find all analytic functions in $D=\{z\in C| |z|<R\}$ such that

$f(2z)=f(3z)$ and $f(0)=1$ for all $z\in D$.

If we take the sequence $z_n=(\frac{2}{3})^n$ then $f$ has the same value in $z_n$, so the sequence $z_n$ converges to $0$ and because of the condition $f(0)=1$ we find that $f(z)=1$ for all $z\in D$.

My question is why $f$ has the same value in $z_n$ ?

Thank you for your help!

Answer 1

In $D$ we can write $f(z)=1+\sum_{n=1}^{\infty}a_nz^n.$ Now there is no reason to think $f(2z),f(3z)$ are also defined everywhere in $D.$ But they are both defined in $D'=\{|z|<R/3\}.$ In $D'$ we then have

$$ f(2z) = 1+\sum_{n=1}^{\infty}a_n2^nz^n =f(3z)=1+\sum_{n=1}^{\infty}a_n3^nz^n.$$

By uniqueness of power series coefficients, we have $a_n2^n=a_n3^n, n=1,2,\dots.$ This implies $a_n=0,n=1,2,\dots,$ hence $f\equiv 1.$

Answer 2

Because by assumption $f(z)=f(3\frac{z}{3})=f(2\frac{z}{3})$ for all $z\in D$. This implies that $f(z_n)=f(z_{n+1})$ for all $n\in\mathbb{N}$.

By the way, just a general note: the term entire function usually means that the function is holomorphic in all $\mathbb{C}$. But maybe there are other definitions as well.