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$1$. 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 $1$

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    $\begingroup$ In 1. $f(z)-z$ has some non-isolated zeros. 2. how did you get that $f$ is constant ? $\endgroup$
    – reuns
    Dec 7 '20 at 3:25
  • $\begingroup$ 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. $\endgroup$
    – MathCosmo
    Dec 7 '20 at 3:29
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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 $z$.

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