the general solution for
$$\ddot{x}(t)+\omega^2 x(t)=0$$
can be expressed as
$$x(t)=Ae^{i\omega t}+Be^{-i\omega t}$$
If $x(t)$ is real then $x(t)=\bar{x}(t)$. That's fine.
But why does it imply that $B=\bar{A}$. I just can't see. Drives me mad.
And please don't tell me because $e^{i\omega t}$ and $e^{-i \omega t}$ are orthonormal. I don't see how. If I picture these two vectors in the complex plane with e.g. $\omega t=30°$ they don't look perpendicular at all. Even if they were...that wouldn't help me anyway.
By constructing the complex conjugate we get $$\begin{align} \overline{x(t)}&=\overline{A\cdot e^{i\omega t}+B\cdot e^{-i\omega t}}\\ &=\overline{A\cdot e^{i\omega t}}+\overline{B\cdot e^{-i\omega t}}\\ &=\overline{A}\cdot e^{-i\omega t}+\overline{B}\cdot e^{i\omega t}\\ &=\overline{B}\cdot e^{i\omega t}+\overline{A}\cdot e^{-i\omega t} \end{align}$$
Where the last equation only equals $x(t)=A\cdot e^{i\omega t}+B\cdot e^{-i\omega t}$ for $A=\overline{B}\Leftrightarrow \overline{A}=B$.