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

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    i mean second,thanks2012-06-22

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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. 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.2012-06-23