Solving Equality Problem in Complex Numbers

[kelvin490]

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?

 

[HallsofIvy]

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 ?

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 e^i2ωt=(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]

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

 

[kelvin490]

I think the concept of linear independence can help. Since e^-iωt and e^iωt 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*)e^i2ωt=(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*.

 

[CellCoree]

I would approach this problem by first clarifying the definitions and assumptions being made. In this case, it seems that x1 and x2 are being defined as equivalent if they have the same values at all times. Additionally, it appears that A and B are being treated as complex numbers with specific properties, such as being conjugates of each other.

Based on these definitions, it is not clear to me how the equation ei2ωt=(A*-B)/(A-B*) would be satisfied if A and B are not equal to each other's conjugate. This is because the numerator and denominator would not have the same values, unless A and B have specific values that satisfy this equation. Therefore, I would say that x1 and x2 cannot be considered equivalent if A and B are not equal to each other's conjugate.

The issue with ei2ωt=(A*-B)/(A-B*)=0/0 being undefined is due to the fact that division by 0 is undefined in mathematics. This problem arises because the equation is assuming that A and B are equal to each other's conjugate, which may not always be the case. If we want to avoid this problem, we would need to be careful in our selection of A and B to ensure that they are not equal to each other's conjugate.