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.

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 $$f(z) = \frac{2}{3+z}.$$

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.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****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.**Click here for more problems. Share to your groups:**

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