# When are the functions $x_1=Ae^{iωt} + Be^{-iωt}$ and $x_2=Ae^{-iωt} + Be^{iωt}$ identically equal?

Suppose we have two equation $x_1=Ae^{iωt} + Be^{-iωt}$ and $x_2=Ae^{-iωt} + Be^{iωt}$. Where $A$ and $B$ are complex number and $A^*$ and $B^*$ are their conjugate correspondingly.

Now if we want to make $x_1$ and $x_2$ exactly equivalent all the time, one way to do it is to have $A=B^*$ and $B=A^*$ so that $x_1$ and $x_2$ 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 $x_1 ≡ x_2$ ?

Another thing trouble me is if $A=B^*$ and $B=A^*$ , then $e^{i2ωt}=(A^*-B)/(A-B^*)=\frac00$ which is undefined. What causes this problem?

• Since $e^{-i\omega t}$ and $e^{i\omega t}$ are linearly independent, $A^*=B$ is the necessary and sufficient condition for $x_1=x_2$. – Math.StackExchange Dec 28 '14 at 5:31

From the context of your equations I infer that $A$ and $B$ are constant coefficients of the exponential functions $e^{-i\omega t}$ and $e^{i\omega t}$. Therefore, an attempt to determine them from the equation $$e^{i2ωt}=\frac{ A^*-B}{A-B^*} \tag{1}$$ is doomed because the left side depends on $t$ while the right side does not. It's impossible to satisfy $(1)$ for all $t$ and have $A,B$ independent of $t$.

One has to be wary of division by something that might be zero. (This was the source of your confusion at the end of the question.) Starting from $(A-B^*)e^{iωt}=(A^*-B)e^{-iωt}$, it is safer to multiply both sides by $e^{iωt}$, thus obtaining $$(A-B^*)e^{2i\omega t}= A^*-B \tag{2}$$ The right hand side of (2) is constant. The left hand side is a periodic complex wave. The only way in which they can equal for all $t$ is for the wave to have amplitude $0$, so that it's always equal to $0$. This means $A=B^*$ (hence $A^*=B$). There are no other solutions.