### CSIR JUNE 2011 PART C QUESTION 79 SOLUTION (Analytic functions and Identity theorem application)

Let $\Bbb D = \{z \in \Bbb C : |z| < 1\}$ be the unit disc. Let $f : \Bbb D \to \Bbb C$ be an analytic function satisfying $$f(\frac{1}{n}) = \frac{2n}{3n+1}.$$ Then
1. $f(0) = \frac{2}{3}$,
2. $f$ has simple pole at $z = -3$,
3. $f(3) = \frac{1}{3}$,
4. no such $f$ exists.

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Solution: We will explicitly calculate the function $f$ then the options are easy to verify.
Identity Theorem: Let $\Bbb D = \{z \in \Bbb C : |z| < 1\}$ be the unit disc. Let $f,g : \Bbb D \to \Bbb C$ be two analytic function. If  $f(\frac{1}{n}) = g(\frac{1}{n})$ for all $n \in \Bbb N$ then $f \equiv g$ (i.e., $f(z) = g(z)$ for all $z \in \Bbb D$).

Replace $n$ by $\frac{1}{n}$ in $\frac{2n}{3n+1}$ we get $\frac{2}{3 + n}$. Let $g : \Bbb D \to \Bbb C$ be an analytic function defined by $g(z) = \frac{2}{3+z}$ (Note that $g$ has no singularities in $\Bbb D$). Then it is given in the problem that  $f(\frac{1}{n}) = g(\frac{1}{n})$ for all $n \in \Bbb N$. Therefore by the above said identity theorem we have $f(z) = g(z)$ for all $z \in \Bbb D$. Hence $$f(z) = \frac{2}{3+z}.$$
option 1: (True) Clearly $f(0) = \frac{2}{3}$.
option 2: (True) Clearly $-3$ is a simple pole of $f(z)$.
option 3: (True) Clearly $f(3) = \frac{1}{3}$
option 4: (False)  The function $f(z) = \frac{2}{3+z}$ is one such function.
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