Sachchidanand Prasad

31-01-2023

Problem: Let $D=\{z\in \mathbb{C} :\vert z \vert < 1\}$ be the unit disc. Let $f:D\to \mathbb{C} $ be an analytic function satisfying \[ f \left( \frac{1}{n} \right) = \frac{2n}{3n+1}. \] Then find out the value of $f(0)+f^\prime (0)$.

Solution: In order to solve this problem, we recall the identity theorem.

(Identity Theorem) Let $\Omega $ be a domain (open and connected) in $\mathbb{C} $ and $f,g:\Omega \to \mathbb{C} $ be analytic functions. If the set $E= \{z\in \Omega :f(z)=g(z)\}$ contains a limit point, then $f(z) = g(z)$ for all $z\in \Omega $.

The above statement sometimes written as

Let $\Omega $ be a domain (open and connected) in $\mathbb{C} $ and $f,g:\Omega \to \mathbb{C} $ be analytic functions. If there exists an infinite sequence $\left\{ z_n \right\} \subset \Omega $ such that $f \left( z_n \right) = g\left( z_n \right),~\forall ~n\in \mathbb{N}$ and $z_n \to z_0\in \Omega $ then $f(z) = g(z)$ for all $z\in \omega $.

Here we define \[ g: D \to \mathbb{C} ,~ z \mapsto \frac{2}{3+z}. \] Note that, \[ g\left( \frac{1}{n} \right) = \frac{2}{3+\frac{1}{n}} = \frac{2n}{1+3n}, \] and $\frac{1}{n}\to 0\in D$. Thus using the identity theorem we conclude that \[ f(z) = g(z),~\forall~z\in D. \] Theretofore, \[ f(z) = \frac{2}{3+z},~\forall~z\in D. \] So, \begin{align*} f(0) + f^\prime (0) = \frac{2}{3} - \frac{2}{9} = \frac{4}{9}. \end{align*}