# The complex conjugate of a complex function.

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 $-i$)

• 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 $z$. – complexmanifold Oct 10 '17 at 13:10
• Yes you are correct. That is the complex conjugate. – SchrodingersCat Oct 10 '17 at 13:10

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