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Complex Numbers

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data
    If [itex]\frac{z_1-2z_2}{2-z_1\overline{z}_2}[/itex] is unimodulus and z2 is not unimodulus, then find |z1|.


    2. Relevant equations



    3. The attempt at a solution
    I am a complete dumb at Complex numbers, please someone guide me in the right direction.
    In this question, what i understand is this, and nothing else.
    [tex]|\frac{z_1-2z_2}{2-z_1\overline{z}_2}|=1[/tex]
    and
    [tex]|z_2|≠1[/tex]
     
    Last edited: Oct 28, 2011
  2. jcsd
  3. Oct 28, 2011 #2

    vela

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    It should be [itex]|z_2| \ne 1[/itex] and
    [tex]\left\lvert \frac{z_1-2z_2}{2-z_1\overline{z}_2} \right\rvert = 1[/tex]Square both sides and use the fact that [itex]|z|^2 = z\bar z[/itex].

    EDIT: Also |w/z| = 1 means |w|=|z|.
     
  4. Oct 28, 2011 #3
    Thanks for the reply vela! :smile:

    I squared both the sides and using the fact [itex]|z|^2 = z\bar z[/itex], i get:-
    [tex]|z_1|^2+4|z_2|^2=4+|z_1|^2|z_2|^2[/tex]

    But now i am stuck here. :(
     
  5. Oct 29, 2011 #4

    Simon Bridge

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    :) Follow your nose: if it was

    x + 4y = 4 + xy

    and you had to find x, what would you do?
     
  6. Oct 29, 2011 #5
    I still dont understand. :(

    Can you give me one more hint? :)
     
  7. Oct 29, 2011 #6
    Thank you both for the help. I have figured it out. :)

    x+4y=4+xy
    or x-xy=4-4y
    or x(1-y)=4(1-y)
    or x=4
    or |z1|=2.

    Thanks again. :)
     
  8. Oct 29, 2011 #7

    Simon Bridge

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    Well done!

    For dessert:

    [tex]z = \left ( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \right )^{12}[/tex]
    ...rewrite z in it's simplest form (it will be exact).
     
  9. Oct 29, 2011 #8
    z=-1. :)
     
  10. Oct 29, 2011 #9

    Simon Bridge

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    Sweet dessert: That one is usually nasty because everyone tries to brute-force it. :)
     
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