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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: $2$

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    @jonas Thanks for your help sir. Ya our analytic functions assumed to be defined on the entire plane.2012-05-13

1 Answers 1

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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

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    Thanks for your answer. earlier i had solved this question.2012-06-12