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Find real x and y, for which |z+3|=1-iz

  1. Dec 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Find real x and y, for which |z+3|=1-iz

    2. Relevant equations
    z=x+iy=re=rcos(θ) + rsin(θ)i

    3. The attempt at a solution
    I know that there must exist some x and y which satisfies both of these equations, since the real part of the LHS must be the same as the RHS, and same with the imaginary part. What is tripping me up is that z is on both sides. If there wasn't a z on either side, I would normally just convert each side into it's polar form, and solve for r and θ, thus finding x and y.

    In this case, I've only gotten so far as changing to polar and doing trivial manipulations to both sides:
    re +3 = 1-ire
    re (1+i) = -2
    re = i-1 = (i-1)ei2πn

    Then I found r=√2, θ=2π. So x=√2 and y =0.

    That's where I'm at so far. I ignored the absolute value of the LHS, which I'm sure probably plays a part in the answer, but I don't quite know how. The answer in the back of the book is x=0 and y=4. I have no idea how to get that answer. Any pointers would be greatly appreciated.
     
    Last edited: Dec 13, 2015
  2. jcsd
  3. Dec 13, 2015 #2

    Ray Vickson

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    If you copied the equation correctly, it has an immediate consequence for ##z##. Since the LHS is a non-negarive real, the right-hand-side must be real as well, so ##iz## must be real.
     
  4. Dec 13, 2015 #3

    Svein

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    Remember: [itex]\left\lvert z+3 \right\rvert [/itex] is real and ≥0. This will give you some boundary conditions for z...
     
  5. Dec 13, 2015 #4

    ehild

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    The absolute value can not be ignored. It makes the left hand side of the equation real: The right hand side must be real, too. What does it mean for y?
     
  6. Dec 13, 2015 #5
    Okay so since |z+3| ≥ 0 does that mean I can now interpret the equation as |z+3| = 1+3i → |z| = -2+3i? I don't think I can, since then r = √13.
     
  7. Dec 13, 2015 #6

    Svein

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    No!
     
  8. Dec 13, 2015 #7
    I don't get it then. What am I missing? What other value is z supposed to have?
     
  9. Dec 13, 2015 #8

    SammyS

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    Have you read all the posts?
     
  10. Dec 13, 2015 #9
    Yes. I know now that the absolute value of the LHS makes everything real, and that |z+3| is some non-negative value. Also, since |z+3| is real, iz must also be real. What I don't understand is the direction I'm supposed to go with that information. And I don't know what that means for y. I'm assuming it means y must be real, but that seems obvious given that the question implies that a real y value exists.
     
  11. Dec 13, 2015 #10

    SammyS

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    Isn't z = x + iy ?

    This gives you that if iz is real, then i(x+ iy) = ix + i2y is real. What can you conclude from this?
     
  12. Dec 13, 2015 #11
    Does it mean that the imaginary part of the LHS would be the real y value of the RHS? Since i(x+iy) = ix+y. When I initially started the problem I wrote |x +iy +3| = 1-ix+y, but I really didn't know what I would do with that.
     
  13. Dec 13, 2015 #12

    SammyS

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    No.
    The left hand side is purely real.

    You said (correctly) that iz is real, which means that ix + i2y is purely real. From that you should be able to conclude something important regarding x or y .
     
  14. Dec 13, 2015 #13

    vela

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    You mean ##x##.
     
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