Tags:
1. Dec 27, 2014

kelvin490

Suppose we have two equation x1=Aeiωt + Be-iωt and x2=A*e-iωt + B*eiωt . Where A and B are complex number and A* B* are their conjugate correspondingly.

Now if we want to make x1 and x2 exactly equivalent all the time, one way to do it is to have A=B* and B=A* so that x1 and x2 are equivalent. However, if we don't do it by this approach but instead set (A-B*)eiωt=(A*-B)e-iωt, then we have ei2ωt=(A*-B)/(A-B*). I would like to ask if the A and B chosen can satisfy this criteria (even A≠B* and B≠A*), can we still say that x1 ≡ x2 ?

Another thing trouble me is if A=B* and B=A* , then ei2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?

Last edited: Dec 27, 2014
2. Dec 28, 2014

HallsofIvy

Staff Emeritus
What you are doing is saying that if $Ae^{i\omega t}+ Be^{-i\omega t}= A^*e^{-i\omega t}+ B^*e^{-i\omega t}$ then we must have $e^{2i\omega t}= \frac{A^*- B}{A- B^*}$. However, the right side of that equation is a constant, independent of t. In order for that to be true, the left side must also be a constant- so what you are doing is equivalent to assuming that $\omega= 0$, a very restrictive condition!

If $A= B^*$ then $B$ must equal $A^*$- that is not a distinct condition. And in that case you are saying $x_1= Ae^{i\omega t}+ A*e^{-i\omega t}$ and $x_2= A*e^{-i\omega t}+ Ae^{i\omega t}$, that is, that $x_1= x_2$ for all t. That is exactly what you said you wanted. There is no "problem". If you have two identical equation, say "px= qy" and "px= qy" and subtract them you get 0x= 0y, of course. It would make no sense to try to "solve for y" in that case, you would just get 0/0 as happens here.

3. Dec 28, 2014

mathman

The two orignal expressions are complex conjugates of each other. Therefore to be equal, the imaginary part must be 0.

4. Dec 28, 2014

kelvin490

I think the concept of linear independence can help. Since e-iωt and eiωt are two linearly independent variables, the coefficient of x1 and x2 must be equal so that x1 and x2 are equivalent.

Also from (A-B*)ei2ωt=(A*-B) we can see that the left hand side is a time dependent function (let's say t is time) and the other side is a constant. The only way to make both sides equal all the time is to have A=B* and B=A*.