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Z^2 = z conjugate

  1. Oct 26, 2011 #1
    The problem statement, all variables and given/known data
    Find all complex solutions to
    z^2 = z conjugate
    i.e. (a+bi)^2 = a-bi

    The attempt at a solution
    First attempt:
    factoring out (a+bi)^2 = a-bi
    leads nowhere.

    Second attempt:
    r^2 (cos2v + isin2v) = r (cos-v + isin-v)
    r must be 1.
    2v = -v + 2∏n
    3v = 2∏n
    v= 2∏n/3
    But that isn't the case.
     
  2. jcsd
  3. Oct 26, 2011 #2

    lurflurf

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    You both forgot z=0.

    The trig way form cos(v)=-1/2 we see |v|=2∏/3
     
  4. Oct 26, 2011 #3
    [tex]\left( a+bi \right)^{2}\; =\; a^{2}-b^{2}+2bi[/tex]
    [tex] a^{2}-b^{2}+2bi=a-bi[/tex]
    How do you, leon, go from there to find a?

    I would claim
    [tex] a\neq a^{2}-b^{2} [/tex]
     
  5. Oct 26, 2011 #4

    vela

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    It would help if you multiplied (a+bi)2 out correctly first.
     
  6. Oct 26, 2011 #5
    The polar form would lead to:
    [tex]
    r^2 = r, \; r \ge 0
    [/tex]
    and
    [tex]
    2\theta = -\theta + 2 k \pi, \; k \in \mathbb{Z}
    [/tex]
    What is the solution to these equations?
     
  7. Oct 26, 2011 #6
    Yes, I forgot the a after 2bi. But that is still not equal to a-bi, which someone posted (seems like that post is now removed)

    I wrote:
    Second attempt:
    r^2 (cos2v + isin2v) = r (cos-v + isin-v)
    r must be 1.
    2v = -v + 2∏n
    3v = 2∏n
    v= 2∏n/3
    But that isn't the case.
     
  8. Oct 26, 2011 #7

    vela

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    [itex]a=a^2-b^2[/itex] is true if you require that [itex]z^2 = \bar{z}[/itex]. That's the point of the problem, to find the values of a and b so that the relationships hold.
     
  9. Oct 26, 2011 #8
    How about writing it as:

    [tex]r^2 e^{2it}=re^{-it}[/tex]

    re-aranage, I get:

    [tex]r=e^{-3it}[/tex]

    Now, the right side is the unit circle. When on the unit circle, is that going to be a positive real number?
     
  10. Oct 26, 2011 #9
    r is not necessarily 1. Also, n can be {0, 1, 2}, because everything else, when taken as an argument of a trigonometric function, gives one of these cases. Also:

    [tex]
    \cos{0} + i \, \sin{0} = 1
    [/tex]

    [tex]
    \cos{\left( \frac{2 \pi}{3} \right)} + i \sin{\left( \frac{2 \pi}{3} \right)} = -\cos{\left( \frac{\pi}{3} \right)} + i \sin{\left( \frac{\pi}{3} \right)} = \frac{-1 + i \sqrt{3}}{2}
    [/tex]

    [tex]
    \cos{\left( \frac{4 \pi}{3} \right)} + i \sin{\left( \frac{4 \pi}{3} \right)} = -\cos{\left( \frac{\pi}{3} \right)} - i \sin{\left( \frac{\pi}{3} \right)} = \frac{-1 - i \sqrt{3}}{2}
    [/tex]
     
  11. Oct 27, 2011 #10
    r doesn't neccessarily have to be 1 but it needs to be a real positive number. this gives you a constraint for t. now by taking r exp(it), you have your solutions.
     
  12. Oct 27, 2011 #11
    Not true, it needs to be a non-negative number.
     
  13. Oct 27, 2011 #12
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