I want to show the following claim is true:
Suppose $f(z)= \Sigma_{n=0}^\infty a_nz^n$ converges in $B_r(0)={\{z\in \mathbb{C} : |z|<r}\}$ for $r>0$, suppose further that if $|2z|< r$ then $f(2z)=(f(z))^2$, and $a_0\neq 0$, then $f(z)=exp(a_1 z)$.
What I would like to do is use uniqueness of a power series: Recall $exp(a_1 z)=\Sigma_{n=0}^\infty \frac{(a_1 z)^n}{n!}$ and we assume $f(z)=\Sigma_{n=0}^\infty a_nz^n$. Fix $z\in B_{r/2}(0)$ and without loss of generality pick a sequence of points ${\{z_k}\}\subset B_{r/2}(0)\setminus {\{(0)}\}$ converging to $z$. My idea was to show that we must have $f(z_k)=exp(a_1 z_k)$ for each $k \geq 1$, which will force the coefficients of the two series to be the same, and thus the series will be equal. But I'm struggling with the computation of showing that $f(z_k)=exp(a_1 z_k)$. Does anyone have any advice on how I could show this?
