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This question is a homework question for my Complex analysis class, and is asked in context of the Identity Theorem (I do not need to show this):

If a holomorphic function defined on a connected open set $U$ is zero at every point of a sequence $\{z_j\}\subset U -􀀀 \{z_0\}$ converging to $z_0 \in U$ as $j$ goes to $1$, then $f$ is identically zero.

Am I not able to conclude that $f(z) = z^2$, and therefore $f(-i) = -1$? The reason I am questioning this is because of the context in which the problem was presented to me, as well as hints from my teacher that this is a "trick question".

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Yes, $f(z)=z^2$. Since $f(1/\sqrt{n})=1/n=(1/\sqrt{n})^2$ for all $n$ and $\lim_{n\to \infty}1/\sqrt{n}=0$, by analytic continuation, we conclude that $f(z)=z^2$. So $f(-i)=-1$.

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  • $\begingroup$ Thank you. My teacher keeps referring to this as a "tricky" question, and I can't for the life of me figure out why. $\endgroup$ – Kyle Funk Nov 16 '15 at 1:00
  • $\begingroup$ The trick is that for analytic function, if it holds for a set with limit points, then it holds for entire domain. $\endgroup$ – Math Wizard Nov 16 '15 at 1:02
  • $\begingroup$ Okay, that makes sense. It seems I was over thinking the problem. $\endgroup$ – Kyle Funk Nov 16 '15 at 1:06

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