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Prove that there is no non-constant analytic function $f$ on $\mathbb{C}$ such that $f(z)=0$ for all $z \in \mathbb{R}$.

Doesn't this follow immediately by the identity theorem? There is a sequence along the real axis that converges to 0...

Attempt

Suppose $f$ is entire and $f(z)=0$ for all real numbers. Now let $E=\mathbb{D}$. Since the set of points of $f=0$ has a limit point in $E$ we conclude $f$ is identically 0.

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  • $\begingroup$ It's a qualifier question and it seems to be resolved by saying there is a sequence so identity theorem. It feels too easy. $\endgroup$ – Aaron Zolotor Jan 8 '18 at 1:44
  • $\begingroup$ I'm trying to figure out exactly how I should phrase it. $\endgroup$ – Aaron Zolotor Jan 8 '18 at 1:45
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    $\begingroup$ Then try to write up the proof. $\endgroup$ – Winther Jan 8 '18 at 1:46
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Indeed “this follow[s] immediately by the identity theorem”. Just take, for instance $(\forall n\in\mathbb{N}):a_n=\frac1n$. Then $\lim_{n\in\mathbb N}=0$ and $(\forall n\in\mathbb{N}):f(a_n)=0$. Therefore, by the identity theorem, $f\equiv0$.

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