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Why $$\sin{ (z)} =\frac{e^{iz}-e^{-iz}}{2i}$$ the only analytic function, is equal to $\sin{(x)}$ for $z=x \in \mathbb{R}$?

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closed as off-topic by José Carlos Santos, Eevee Trainer, Xander Henderson, Lee David Chung Lin, mrtaurho Feb 20 at 6:31

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  • $\begingroup$ Why are you interested in this question and what have you tried so far? $\endgroup$ – James Feb 19 at 14:42
  • $\begingroup$ It's a task, that I found on the internet. I have no idea, how I can argue here. Do you have a hint? $\endgroup$ – Leon1998 Feb 19 at 14:50
  • $\begingroup$ Try the identity theorem from complex analysis. $\endgroup$ – James Feb 19 at 14:53
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Lemma Let $f,g$ be two entire functions which are equal on $\mathbb R$. Then, they are equal on $\mathbb C$.

Proof Since $f-g$ is analytic and the set of zeroes has an acumulation point, by the identity theorem for analytic functions $f-g=0$.


Now apply this Lemma to $f(z)=\sin(z)$ and $g(z)=\frac{e^{iz}-e^{-iz}}{2i}$.

Alternatelly, you can argue that any function which is equal to $\sin(z)$ on $\mathbb R$ must be analytic on $\mathbb R$, and hence its Taylor series is the Taylor series of $\sin(z)$. But by Taylor Theorem, any such function must be equal to the Taylor series on $\mathbb C$.

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  • $\begingroup$ I have a question to your proof: Why does the set of zeroes have an acumulation point? $\endgroup$ – Leon1998 Feb 19 at 14:59
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    $\begingroup$ @Leon1998 0 for example is an accumulation point, since it is the limit of $\frac{1}{n} \in \mathbb R$ $\endgroup$ – N. S. Feb 19 at 15:10
  • $\begingroup$ Thank you, that means applied for my example: The set$ f(z_0)=g(z_0)$ with $z_0=0$ has an accumulation point in C. Is that ok? $\endgroup$ – Leon1998 Feb 19 at 15:23
  • $\begingroup$ @Leon1998 Do you know what an acumulation point is? $\endgroup$ – N. S. Feb 19 at 15:36
  • $\begingroup$ Yes, what's the problem in my proof. Where is the mistake? $\endgroup$ – Leon1998 Feb 19 at 16:01

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