Prove or Disprove that | e 2 i θ − 2 e i θ − 1 e 2 i θ + 2 e i θ − 1 | = 1This is a step in an attempt to solve a much larger problem, thus I'm fairly sure it's true but not absolutely sure. It looks like it should be simple but it's resisted all my attempts so far.

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Prove or Disprove that       |                            e                      2            i            θ                          −        2                  e                      i            θ                          −        1                              e                      2            i            θ                          +        2                  e                      i            θ                          −        1              |    =  1This is a step in an attempt to solve a much larger problem, thus I&#039;m fairly sure it&#039;s true but not absolutely sure. It looks like it should be simple but it&#039;s resisted all my attempts so far.

Corinne Choi · Accepted Answer

This is true.      |    z  −      1    z    −  2      |    =      |    z  −      1    z    +  2      |  where   z  =      e          i      θ      , since   ℜ  (  z  −      1    z    )  =  0Geometrically,   z  −      1    z   lies on the y-axis (perpendicular bisector of (2,0) and (−2,0)).

mars6svhym · Accepted Answer

Divide by       e          i      θ       the numerator and denominator :      |                            e                      2            i            θ                          −        2                  e                      i            θ                          −        1                              e                      2            i            θ                          +        2                  e                      i            θ                          −        1              |    =      |                            e                      i            θ                          −        2        −                  e                      −            i            θ                                                e                      i            θ                          +        2        −                  e                      −            i            θ                                |  Think at the complex conjuguate of the numerator and conclude!

Prove or Disprove that | e <mrow class="MJX-TeXAtom

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