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What is $\overline{\sin(z)}$ equal to?

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Do you mean $\sin(\bar{z})$, or do you mean $\overline{\sin(z)}$? – Arturo Magidin Jun 22 '12 at 16:22
i mean second,thanks – Arash Jun 22 '12 at 16:23
up vote 4 down vote accepted

I will write $\exp(x)$ instead of $e^{x}$, they are synonyms. (Just a notational warning!)

Well, recall Euler's formula $$ \exp(i\theta)=\cos(\theta)+i\cdot\sin(\theta).$$ Then we see $$\exp(i\theta)-\exp(-i\theta)=2i\cdot\sin(\theta)$$ allows us to write $$\frac{\exp(i\theta)-\exp(-i\theta)}{2i}=\sin(\theta).$$ So replacing $\theta$ with $z=x+iy$ in this case would produce $\exp(iz)-\exp(-iz)=2i\cdot\sin(z)$, where $z=x+iy$.

Addendum: Now consider the following: $$\overline{\sin(z)} = \overline{\frac{\exp(iz)-\exp(-iz)}{2i}}$$ But look, this is just $$\overline{\sin(z)}=\overline{\sin(x+iy)}$$ and the only place the imaginary part plays any role is the $iy$, we have $$\overline{\sin(z)}=\sin(x-iy)$$ and this is precisely $\sin(\bar{z})$.

Looking on the right hand side, how can we say this? Well, we just change all the signs for $i$ and replace $z$ with $\bar{z}$, writing $$\overline{\frac{\exp(iz)-\exp(-iz)}{2i}}=\frac{\exp(-i\bar{z})-\exp(i\bar{z})}{-2i}$$ But look, we may multiply the top and bottom by $-1$ producing $$\overline{\frac{\exp(iz)-\exp(-iz)}{2i}}=\frac{\exp(-i\bar{z})-\exp(i\bar{z})}{-2i}=\frac{-\exp(-i\bar{z})+\exp(i\bar{z})}{2i}$$ which is precisely $\sin(\bar{z})$.

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thanks,but i mean sin(z) bar(all of sin(z) is under bar not just z) you edited my question mistakenly – Arash Jun 22 '12 at 16:18
@Arash, my error :$ I'm terribly sorry about that; however, I think this should be the correct answer... – Alex Nelson Jun 22 '12 at 16:24
thanks so much,this is correct – Arash Jun 22 '12 at 16:27
@bogus: It is a standard exercise on analytic continuation to show that that is, indeed, the case provided that: A) $f(x)$ is real for all $x$, B) $f$ is holomorphic on a domain that is symmetric about the real axis and intersects it in an open interval. – Jyrki Lahtonen Jun 22 '12 at 22:08
Thanks. I would like to emphasize, for anyone reading your comment, that $f(x)$ must be real for all real $x$, otherwise the answer to my question is no. – Stefan Smith Jun 23 '12 at 22:57

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