- #1

kelvin490

Gold Member

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Suppose we have two equation x

Now if we want to make x

Another thing trouble me is if A=B* and B=A* , then e

_{1}=Ae^{iωt}+ Be^{-iωt}and x_{2}=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 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*)=0/0 which is undefined. What causes this problem?
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