# Analytic continuation of $\sin(z)$ [closed]

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Eevee Trainer, Xander Henderson, Lee David Chung Lin, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.

• Why are you interested in this question and what have you tried so far? – James Feb 19 at 14:42
• It's a task, that I found on the internet. I have no idea, how I can argue here. Do you have a hint? – Leon1998 Feb 19 at 14:50
• Try the identity theorem from complex analysis. – James Feb 19 at 14:53

## 1 Answer

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 $$\mathbb C$$.

• I have a question to your proof: Why does the set of zeroes have an acumulation point? – Leon1998 Feb 19 at 14:59
• @Leon1998 0 for example is an accumulation point, since it is the limit of $\frac{1}{n} \in \mathbb R$ – N. S. Feb 19 at 15:10
• 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? – Leon1998 Feb 19 at 15:23
• @Leon1998 Do you know what an acumulation point is? – N. S. Feb 19 at 15:36
• Yes, what's the problem in my proof. Where is the mistake? – Leon1998 Feb 19 at 16:01