Which one of the following two statements is/are true?
$(\text a)$ There exists an analytic function $f\colon\Bbb C \longrightarrow \Bbb C$ such that $f(z) = z$ for all $z$ with $|z|=1$ and $f(z) = z^2$ for all $z$ with $|z|=2.$
$(\text b)$ There exists an analytic function $f\colon\Bbb C \longrightarrow \Bbb C$ such that $f(0) = 1,f(4i) = i$ and for all $z_j$ with $1 < |z_j| < 3,$ $j = 1,2,$ we have $$|f(z_1) - f(z_2)| \leq |z_1 - z_2|^{\pi/3}.$$
How do I proceed to find the correct option. Any help in this regard will be highly appreciated.
Thank you very much for your valuable time.
a) It follows from that identity theorem that if $\lvert z\rvert=1\implies f(z)=z$, then $f(z)=z$ everywhere. Therefore, the answer is negative.
b) If $1<|z|<3$, then$$f'(z)=\lim_{w\to z}\frac{f(w)-f(z)}{w-z}=0,$$since$$\left|\frac{f(w)-f(z)}{w-z}\right|\leqslant|w-z|^{\pi/3-1}$$and $\frac\pi3-1>0$. So, $f$ is constant on the annulus $\{z\in\Bbb C\mid1<|z|<3\}$. Therefore, again by the identity theorem, it is constant, and therefore we can't have both $f(0)=1$ and $f(4i)=i$.