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Nonzero complex numbers

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider 3 nonzero complex numbers $$z_1,z_2,z_3$$ each satisfying $$z^2=i \bar{z}$$. We are supposed to find $$z_1+z_2+z_3, z_1z_2z_3, z_1z_2+z_2z_3+z_3z_1$$.

    The answers- 0, purely imaginary , purely real respectively.



    2. Relevant equations


    3. The attempt at a solution
    I have no idea how to proceed. I tried to use the expansion for $$(a+b+c)^2$$ for them, but I am not getting anywhere. Please help. Thanks in advanced!
     
  2. jcsd
  3. Sep 15, 2016 #2

    BvU

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    Asking you to show your work doesn't make sense for this expansion. Did you try to work out what ##z^2 = i\bar z ## means for the real and imaginary parts ##\alpha## and ##\beta## if you write ##z = \alpha + i\beta## ?
     
  4. Sep 15, 2016 #3
    Hi cr7:

    I suggest that you work with the equation
    z2 = i zbar .​
    Try to understand the constraints on values of z that satisfy this equation. I suggest thinking about z in polar form. You also might find the following of some help:

    Good luck.

    Regards,
    Buzz

    PS. I have no idea why that PNG file got attached. Please ignore it.
     

    Attached Files:

  5. Sep 15, 2016 #4

    BvU

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    No typos in the problem statement ?
     
  6. Sep 15, 2016 #5

    haruspex

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    Do those three expressions remind you of any standard formulas?
     
  7. Sep 16, 2016 #6
    Yes @haruspex I got the answer .....the relation between coeffecients and roots....
     
  8. Sep 16, 2016 #7

    PeroK

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    Two things. The question ought to say that ##z_1, z_2, z_3## are distinct. Also, can you not get much a much more specific answer than that?
     
  9. Sep 20, 2016 #8

    BvU

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    Can someone help me out ? If I follow my own advice (#2) and solve, I get pure imaginary, 0, pure real, respectively (hence my #4, and indeed, the three roots must be distinct to get that).
     
  10. Sep 20, 2016 #9

    PeroK

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    This was definitely one to use polar form.
     
  11. Sep 20, 2016 #10

    BvU

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    Same thing. Still I get an imaginary number as sum, zero as a product and one real nonzero term in sum of dyadic products ... :oldconfused:
     
  12. Sep 20, 2016 #11

    PeroK

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    If ##z^2 = i \bar{z}## then, first, we get ##|z| = 1## and then ##z^3 = i##.
     
  13. Sep 20, 2016 #12

    BvU

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    So ##\cos{\pi\over 6} + i\sin {\pi\over 6},\ \ 0,\ \ \cos{5\pi\over 6} + i\sin {5\pi\over 6}##.
    Sum imaginary, product zero, and sum of dyadic products real, respectively ... :oldconfused:
     
  14. Sep 20, 2016 #13

    PeroK

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    I just read off the coefficients of ##z^3 - i = (z-z_1)(z-z_2)(z-z_3)##

    ##z = 0## is a fourth solution, but the problem states that the ##z_i## are non-zero. There's your problem!
     
  15. Sep 20, 2016 #14

    Ray Vickson

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    No: from ##z^2 = i \bar{z}## it follows that either ##z=0## or else ##|z| = 1## and then ##z^3 = i##. There are three nonzero roots of ##z^3 = i##.
     
    Last edited: Sep 20, 2016
  16. Sep 20, 2016 #15

    BvU

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    Finally got it ..
    o:) Thanks guys !
     
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