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Does there exists an analytic function $f$ in unit disk such that $f(-\frac{1}{2})=3$, $f(n^{-2})=5$ for $n\ge 2$. i am not able to solve any help will be appreciated.

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By continuity, $f(0)=5$. But then the zero at $z=0$ of the function $f(z)-5$ is not isolated. We conclude $f(z)-5=0$

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Suppose that , $f(z)$ is analytic in the unit disc.

Consider $g(z)=f(z)-5$. Then $g\left(\frac{1}{n^2}\right)=f\left(\frac{1}{n^2}\right)-5=0$.

So, the set of all zeros of $f(z)$ , $\left\{\frac{1}{n^2}:n\in \mathbb N\right\}$ has a limit point $0$ in the unit disc.

So, by Identity theorem $g$ is identically $0$. Then $f(z)=5$.

Now, $f(-1/2)=5$, which as a contradiction.

Hence there are no such function satisfying the given conditions.

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