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

[Niles]

Homework Statement

Hi

Say I have the following equation:

<br /> Ae^{ix}+Be^{-ix} = Ce^{ix}+De^{-ix}<br />

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

Ae^{ix} = Ce^{ix} and <br /> Be^{-ix} = De^{-ix}<br />

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}<br /> A+B=C+D\\<br /> A-B=C-D<br /> \end{array}\right.

This is easily solved...