Problem about equivalence

[kelsiu]

Suppose we have two equation x₁=Ae^iωt + Be^-iωt and x₂=A*e^-iωt + B*e^iωt . 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*)e^iωt=(A*-B)e^-iωt, then we have e^i2ωt=(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 e^i2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?

 

[Ishuda]

Suppose we have two equation x₁=Ae^iωt + Be^-iωt and x₂=A*e^-iωt + B*e^iωt . 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*)e^iωt=(A*-B)e^-iωt, then we have e^i2ωt=(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 e^i2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?

The problem comes in the step from (A-B*)e^iωt=(A*-B)e^-iωt [ (A-B*)e^2iωt=(A*-B) ] to e^i2ωt=(A*-B)/(A-B*). You can only perform this step if (A-B*) is not zero. It has the same fallacy as the 'proof' that all integers are equal to each other: Let
x = y
then
1 = y/x.
Now if we let y = 0 we see that
1 = 0
and adding 1 to each side we have
2 = 1 = 0,
etc.