Let $S = \{0\}\cup \{\frac{1}{4n+7} : n =1,2\ldots\}$. How to find the number of analytic functions which vanish only on $S$?
Options are
a: $\infty$
b: $0$
c: $1$
d: $2$
Let $S = \{0\}\cup \{\frac{1}{4n+7} : n =1,2\ldots\}$. How to find the number of analytic functions which vanish only on $S$?
Options are
a: $\infty$
b: $0$
c: $1$
d: $2$
First I would like to say that zeroes of analytic function are isolated point.Identity Theorem or In some books uniqueness theorem says that: $f$ be analytic in a domain $D$, If the set of zeroes has a limit point in the domain $D$ then $f\equiv 0$. In your case $D=\mathbb{C}$ and set of zeroes=$S$(as you have already defined in your question),Notice that $S$ has a limit point namely $0\in S$, so Uniqueness theorem says that only all the analytic function that has zero set as $S$ must be $\equiv 0$ function