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

[Niles]

Homework Statement

Hi

Say I have the following equation:

$$Ae^{ix}+Be^{-ix} = Ce^{ix}+De^{-ix}$$

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

$Ae^{ix} = Ce^{ix}$ and $Be^{-ix} = De^{-ix}$

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

 

[micromass]

The equation $$Ae^{ix}+Be^{-ix}=Ce^{ix}+De^{-ix}$$. This yields

$$(A+B)\cos(x)+(A-B)i\sin(x)=(C+D)\cos(x)+(C-D)i\sin(x)$$

Thus this gives us a system of equations:

$$\left\{\begin{array}{c} A+B=C+D\\ A-B=C-D \end{array}\right.$$

This is easily solved...