All entire functions which satisfying : $f(2z)=f(z)^{2}$ I want to solve following question:   

Find all entire functions satisfying the condition that $f(2z)=f(z)^{2}$ and $f(0)\ne 0.$   

I know the function $f(z)=\mathbb e^{wz}$ for every $w$ is a solution. Is there another entire function by above property?
 A: First, it is easy to see that
$$f(z) \ne 0 \text{ for all } z \in \Bbb C \, .$$
(Assume that $f(z_0) = 0$. It follows that $f(z_0/2) = 0$ and therefore
$f(z_0/2^k) = 0$ for all $k \in \Bbb N$. Then the zeros of $f$ acccumulate
at $z=0$, and it follows from the Identity theorem that $f$ is identically
zero.)
So $g(z) := f'(z)/f(z)$ is an entire function.
Differentiating $f(2z)=f(z)^{2}$ gives
$$
 2f'(2z) = 2f(z)f'(z) \, .
$$
and therefore
$$
  g(2z) = \frac{f'(2z)}{f(2z)} = \frac{f(z)f'(z)}{f(z)^2} = g(z) \, .
$$
Now we can argue as above: For an arbitrary $z_1 \in \Bbb C$, $C := g(z_1) = g(z_1/2^k)$ for all $k \in \Bbb N$, and it follows from the identity theorem that $g \equiv C$.
So we have $f'(z) = Cf(z)$ for all $z \in \Bbb C$, and this implies
$f(z) = De^{Cz}$  for some $D \in \Bbb C $.
(If that isn't obvious, define $h(z) := f(z) e^{-Cz}$.
Then $h' \equiv 0$ so that $h$ is constant.) 
Finally we can substitute $f(z) = De^{Cz}$ into $f(2z)=f(z)^{2}$
and conclude that $D = 1$, so that all solutions are given by
$$
f(z) = e^{Cz} \,, \quad C \in \Bbb C \, .
$$
A: We clime that $f(z)\neq0$ for every $z\in \mathbb C$. If $f(z)=0$ for some $z\in \mathbb C$. by the hypothesis, $(f(\frac{z}{2}))^2=f(z)=0$. So $f(\frac{z}{2})=0$. By induction, $f(\frac{z}{2^k})=0$. When $k\to \infty$, $\frac{z}{2^k}\to 0$. Since $f$ is entire, so $f(0)=0$, contradict. So, $f(z)\neq0$ for every $z\in \mathbb C$. From $f(2z)=f(z)^2$, we can get: $f(0)=1$, and: $f'(2z)=f'(z)f(z)$. If for some $z\neq 0$, $f'(z)=0$, then, for every $k\in \mathbb N$, we should have: $f'(\frac{z}{2^k})=0$ and when $k\to \infty$, we should have: $f'(0)=0$. But $f'$ is entire and the set $\{z\in \mathbb C; f'(z)=0\}$ has a limit point. So $f'(z)=0$ for every $z\in \mathbb C$. its mean $f(z)=1$ for every $z\in \mathbb C$.
Now let $f'(0)\neq 0$. define $g(z)=\frac{f(z)}{\mathbb e^{f'(0)z}}$. Then, it is easy to check that:
$$
g(2z)=(g(z))^2
\,\,\,\,\,\, (1)$$
We know: $g,g'$ are entire and $\,\,$ $g(0)=1$ $\,,\,$ $g'(0)=0$.
If we prove: $g^{(n)}(0)=0$ for every $n\in \mathbb N$, then $g\equiv 1$ and we are done.
Consider that $g$ is entire. So $g$ has a power series $$g(z)=a_0+a_1z+a_2z^2+a_3z^3+...+a_nz^n+...$$ which for every $z\in \mathbb C$ is convergence and $a_n=\frac{g^{(n)}(0)}{n!}$. From $g(0)=1$ $\,,\,$ $g'(0)=0$ we get: $$g(z)=1+a_2z^2+a_3z^3+...+a_nz^n+...$$ From equation $(1)$ we should have: $$
\begin{align}
1+4a_2z^2+8a_3z^3+...+2^na_nz^n+...&=(1+a_2z^2+a_3z^3+...+a_nz^n+...)^2\\
&=1+c_2z^2+c_3z^3+...+c_nz^n+...
\end{align}
$$ Which $c_n=2a_n+a_2a_{n-2}+a_3a_{n-3}+...+a_{n-2}a_2$. From here: $4a_2=2a_2$ which mean: $a_2=0$. By induction, let $a_1=0,a_2=0,...,a_n=0$. We want to prove: $a_{n+1}=0$. $$2^{n+1}a_{n+1}=2a_{n+1}+\sum_{j=1}^{n-1}a_ja_{n-j}$$
But $\sum_{j=1}^{n-1}a_ja_{n-j}=0$, and it's mean $a_{n+1}=0$. So $g\equiv 1$. Thus: $f(z)=\mathbb e^{f'(0)z}$ are all of function's satisfying $f(2z)=(f(z))^2$
A: Hint: Let $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Then 
$$f(2z) = \sum_{n=0}^{\infty}(2^na_n)z^n,\,\,\, \, f(z)^2 = \sum_{n=0}^{\infty}\left (\sum_{k=0}^{n}a_ka_{n-k}\right )z^n.$$
Equate coefficients for small $n$ and a nice pattern emerges, which you can then prove by induction.
