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Complex number equality problem

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem states that you need to solve the following equation (without a calculator) : z^5 = z̅

    2. Relevant equations
    z=a+bi and z̅=a-bi

    3. The attempt at a solution
    So far I've tried multiplying both sides by z̅: z̅ * z^5 = |z̅|^2
    (is z̅ * z^5 = z^6 ??) |z|^6 = |z̅|^2

    I then divided both sides by |z|^2, but this has lead to an incorrect answer. The answer given is z^6=1 ,
    which then leads to a root of unity equation.

    Thank you in advance.
     
  2. jcsd
  3. Nov 18, 2015 #2

    Samy_A

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    No, or more correctly, not necessarily.

    What happens when you multiply both sides by z?
    (Note that ##z\bar z
    = |z|²##)
     
  4. Nov 18, 2015 #3

    Ray Vickson

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    Have you tried the polar representation ##z = r e^{i \theta}##?
     
  5. Nov 19, 2015 #4
    Hi
    I have tried r5( cos(5 theta) + i sin(5 theta) ) = r( cos(theta) +i sin(theta) ) but this lead to:
    cos(5 theta) = cos (- theta) and i sin(5 theta) = i sin(-theta)
    which is very difficult to solve without a calculator. Using the polar form results in similarly difficult equations.
    Thanks
     
  6. Nov 19, 2015 #5
    The polar form results in : (reitheta)6=re-itheta
    e6itheta=e-i theta
    e7itheta=0
    cos(7theta) + isin(7theta)=0
     
  7. Nov 19, 2015 #6

    Samy_A

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    The 6 should be 5.
    This is wrong
    If ##e^{ix}=e^{iy}##, then ##e^{i(x-y)}=1##

    Actually, wether you use my hint, or Ray's, you should easily get to the root of unity equation you expect.
     
    Last edited: Nov 19, 2015
  8. Nov 19, 2015 #7

    Ray Vickson

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    The polar form give conditions on both ##r## and ##\theta##, but you have said nothing about ##r##.
     
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