Gold Member
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?

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HallsofIvy
Homework Helper
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 ?
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!

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?
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.

mathman