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Does there exist analytic function with $f(\frac{i^n}{n})=\frac{-1}{n^2} \forall n\ge 2$, well I guess Yes, beacuse $g(z)=f(z)+z^2$ has zero set $\{-\frac{i}{n}: n \text{ odd}\}$ which has limit point zero, hence $f(z)=-z^2$ Is my answer is correct?

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If you plug $z = i^n/n $ into $f(z) = -z^2$, you don't always get $-{1 \over n^2}$. Therefore... –  Zarrax Jun 10 '12 at 21:19
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2 Answers

up vote 2 down vote accepted

If you plug in $i^3/3$, you'll see that $f(z)=-z^2$ fails to satisfy the desired property.

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Your reasoning is basically correct but your answer is not. You have deduced that the only possible solution is $-z^2$, but even this fails, so there is no solution.

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