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Hi I'm trying to work out the complex conjugate of: $Ae^{(-a(mx+it))}$. Generally when I have tried to work out a complex conjugate of a complex number I just replace $i$ with $-i$. However I didn't know if it would work this time due to this being a complex function of two variables ($x$ and$ t$ who are real). When I do this and multiply the complex conjugate with the complex function I get a purely real answer which makes me think I'm right but like I said I've never done it like this before.

Also if this is right then is this what I can do with every complex function (replace $i$ with $-i$)

Thanks in advance.

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    $\begingroup$ Yes, you are right. let $z=A\exp(-amx-ait)$ then $\bar{z}=A\exp(-amx+ait)$ providing $A$ is real of course. The real answer you refer to when you find $z \bar{z}$ is just the square of the modulus of $z$. $\endgroup$ – complexmanifold Oct 10 '17 at 13:10
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    $\begingroup$ Yes you are correct. That is the complex conjugate. $\endgroup$ – SchrodingersCat Oct 10 '17 at 13:10
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For simplicity, I assume that $A = a = 1$. In general, for real $t$, $$ \overline{e^{it}} = \overline{\cos t + i\sin t} = \cos t - i\sin t = \cos(-t) + i\sin(-t) = e^{-it}. $$ Hence, for your expression you get $$ \overline{e^{mx+it}} = \overline{e^{mx}e^{it}} = e^{mx}e^{-it} = e^{mx-it}. $$ Hope this helps for understanding. If something is unclear, do not hesitate to ask in the comments.

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Hint:

$$\overline{r(\cos(\theta)+i\sin(\theta))}=r\cos(\theta)-ir\sin(\theta)=r(\cos(-\theta)+i\sin(-\theta)).$$

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – kjetil b halvorsen Oct 10 '17 at 14:21
  • $\begingroup$ @kjetilbhalvorsen: that's clearly tagged as a hint. It is actually very close to a complete solution, and I have made it even closer. $\endgroup$ – Yves Daoust Oct 10 '17 at 16:01

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