Finding the number of analytic functions which vanish only on a given set.

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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Hint: the zeroes of an analytic function can't be arbitrarily close together. – Pedro May 13 '12 at 7:09
@Pedro Then what would be conclusion sir? – srijan May 13 '12 at 7:16
@pedro Can you write the answer with little explanation please? I would be very much thankful to you. – srijan May 13 '12 at 7:32
Actually the zeros of an analytic function can be arbitrarily close together, e.g. $\sin(z^2)$. But the zero set an analytic function whose domain is a connected open subset of the plane cannot have an accumulation point in the domain, unless the function is identically zero. – Jonas Meyer May 13 '12 at 7:49
@srijan: Have you seen the "identity theorem"? (If you scroll down to "an improvement" you will find stuff about accumulation points.) That article refers to $2$ analytic functions $f$ and $g$, but the version Pedro and I are alluding to is the special case where $g=0$. So suppose that you have an analytic function $f$ that is zero on $S$. Can you find a way to apply the identity theorem to make a conclusion about $f$? By the way, are your analytic functions assumed to be defined on the entire plane? – Jonas Meyer May 13 '12 at 8:07

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