Prove/disprove the existence of an analytic map using Identity Theorem

Question

I want to prove/disprove the following statement using Identity Theorem.

Does there exists an analytic map $f: \Bbb C \to \Bbb C$ such that $f(z)=z$ for all $z$ such that $|z|=1$ and $f(z)=z^2$ for all $z$ such that $|z|=2$

Is there any idea where I can start?

Any help is appreciated. Thank you.

Answer 1

Since $f(z)=z$ when $\lvert z\rvert=1$ and since the set $\{z\in\mathbb C\,|\,\lvert z\rvert=1\}$ has non isolated points (actually, it has no isolated point), it follows from the identity theorem that $(\forall z\in\mathbb C):f(z)=z$. Therefore, $f(2)=2\neq2^2$, and so there is no such function.

Answer 2

Consider the function $g(z)=f(z)-z.$ Note that $g$ is analytic on all of $\Bbb C$ and is identically $0$ on the unit circle. What does this let you conclude about $g$? What does this let you conclude about $f$?