Suppose we have two equation x(adsbygoogle = window.adsbygoogle || []).push({}); _{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?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Problem about equality

Tags:

Loading...

Similar Threads - Problem equality | Date |
---|---|

I A problem in combinatorics | Jan 17, 2018 |

B Optimisation problem | Jan 11, 2018 |

I Math papers and open problems | Dec 11, 2017 |

I Proof: 0.9999 does not equal 1 | Nov 22, 2017 |

**Physics Forums - The Fusion of Science and Community**