Existence of analytic function with certain conditions

$$1$$. Does there exists an analytic function $$f: \mathbb{C} \rightarrow \mathbb{C}$$ such that $$f(z)=z$$ for all $$|z|=1$$ and $$f(z)=z^2$$ for all $$|z|=2$$.

$$2$$.There exists an analytic function $$f: \mathbb{C} \rightarrow \mathbb{C}$$ such that $$f(0)=1,f(4i)=i$$ and for all $$z_j$$ such that $$1 < |z_j| < 3, j=1,2$$ we have $$|f(z_1)-f(z_2)| \leq |z_1-z_2|^{\frac{\pi}{3}}$$

For $$2$$, I get that $$f$$ is constant on the annulus. Then by identity theorem, I can conclude that f cannot exist. Am I right?

I don't know how to proceed $$1$$

• In 1. $f(z)-z$ has some non-isolated zeros. 2. how did you get that $f$ is constant ? Dec 7 '20 at 3:25
• I divide both sides with |z1-z2| then taking z1 tends to z2 we get that the derivative is 0. This is true for all z in that annulus right? Then Identity theorem is forcing f to be constant for the whole C. Dec 7 '20 at 3:29

Your answer for second part is correct. The first part is much simpler: $$f(z)-z$$ is analytic on $$\mathbb C$$ and its zeros have a limit point. Hence $$f(z)=z$$ for all $$z$$.