let $ D=$ {$z\in\mathbb C:|z-1|<1 $}. Consider the analytic function $f(z)$ on $D$ such that $f(1)=1$ and $f(z)=f(z^2)$ for all $z\in D$.Then which of the following is not true?
1) $f(z) = [f(z)]^2$ for all $z\in D$
2) $f(\frac{z}{2}) = \frac{1}{2}f(z)$ for all $z\in D$
3) $f(z^3) = [f(z)]^3$ for all $z\in D$
I've taken $f(z)=1$. Hence option 2) is the answer. Is it true? Is there any other method to solve the problem because I don't know how to prove options 1) and 3) ? Please Help...
This question is there in MSE like here , here!
Still I would like to add that all boils down to know that $f$ is constant (how?)! Like consider a sequence $z_{n}$ as $z_{0} = \frac{1}{2}$ , $z_{n+1} = \sqrt{z_{n}}$ so $f(z_{n+1}) = f(z_{n}) \forall n \in \Bbb{N}$ and $z_{n} \rightarrow 1$ and now as $f$ is continuous so $f(z_{n}) \rightarrow f(1) = 1$, so all the values of $f(z_{n} ) = 1 \forall n \in \Bbb{N}$ now from identity theorem it follows that $f$ is aconstant function so $f(z) = 1 \forall z \in D$ . So clearly when option 1 and 3 are correct and option 2 is wrong!
