Complex Equation Homework: Ae^(ix)=Ce^(ix) & Be^(-ix)=De^(-ix)

[Niles]

Homework Statement

Hi

Say I have the following equation:

[tex]
Ae^{ix}+Be^{-ix} = Ce^{ix}+De^{-ix}
[/tex]

then my book says that the above implies that we have the two equations

[itex]Ae^{ix} = Ce^{ix}[/itex] and [itex]
Be^{-ix} = De^{-ix}
[/itex]

since it must be valid for all x. I cannot see why?Niles.

 

[micromass]

The equation [tex]Ae^{ix}+Be^{-ix}=Ce^{ix}+De^{-ix}[/tex]. This yields

[tex](A+B)\cos(x)+(A-B)i\sin(x)=(C+D)\cos(x)+(C-D)i\sin(x)[/tex]

Thus this gives us a system of equations:

[tex]\left\{\begin{array}{c}
A+B=C+D\\
A-B=C-D
\end{array}\right. [/tex]

This is easily solved...