What is $\overline{\sin(z)}$ equal to?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||
|
|
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 $\sin(\bar{z})$. |
|||||||||||||||
|
