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Complex polynomial solutions

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    (Z^3)+3i(conjugate z) = 0
    find all solutions.

    2. Relevant equations


    3. The attempt at a solution
    How can i isolate Z Tried factoring out z, didn't came out good: z(z^2+3i*(Conjugate z)/z) ==> Right part equal to zero. couldn't factor anymore.
     
  2. jcsd
  3. May 10, 2016 #2

    Samy_A

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    You could use ##z=x+iy##, ##\bar z = x-iy##, and see what equations you get when you plug that in into ##z³+3i\bar z=0##.
     
  4. May 10, 2016 #3
    so i opened it and came out something hairy with a^3 - b^3i -3ab(b-ai) +3(b+ai) = 0
    (a^3)-3(b^2)a+3b+(3(a^2)b-b^3+3a)i ==>
    (a^3)-3(b^2)a+3b = 0
    3(a^2)b-b^3+3a = 0

    What can i try to do here?

    btw the solutions can be shown as trigonometric (cosx+isinx) aswell.
     
    Last edited: May 10, 2016
  5. May 11, 2016 #4

    ehild

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    Write z in trigonometric form z=r(cosx+isinx) into the equation.
     
  6. May 11, 2016 #5
    r3(cos3x+isin3x) = 3ir(cosx + isin(-x))

    ==> r3 = 3ir ==> r = squareroot(3)

    ==> cos3x+isin3x = cosx + isin(-x)
    ==> cos3x = cosx (1)
    ==> isin3x = isin(-x) (2)

    cos2x = 1 (1)
    isin2x = -1 (2)

    which x will satisfy both? or do i have a mistake?
     
    Last edited: May 11, 2016
  7. May 11, 2016 #6

    ehild

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    You can not do that! i can not be ignored.
    It is wrong again. You ignored r.
    cos(3x) is not 3cos(x) and so on.

    r3(cos3x+isin3x) = 3ir(cosx + isin(-x))
    one solution can be that r=0. If r ≠ 0, simplify with r. Then collect the real and imaginary terms. You get two equations, one for the real terms, and one for the imaginary ones.
     
  8. May 11, 2016 #7
    Ok so r^3= 3ir ==> r^2 =3i ==> r = sqrt(3i).
    i just remember that r must be real, since it the size of the vector.
     
  9. May 11, 2016 #8

    ehild

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    Yes, r is real, so r3 can not be equal to ir.
    Let's start again. z = r (cos(θ) + i sin (θ)). You wrote z3 correctly, z3 = r3(cos (3θ) + i sin (3θ) ) and z* = r( cos(-θ)+isin(-θ).
    The original equation is (Z^3)+3i(conjugate z) = 0, which means z3=-iz*. The magnitude of both sides has to be the same . What equation you get for r?
    The phase angles also should be the same, or differ by 2kπ (k integer) . Write the right side as cos(φ)+ isin(φ) using that -i corresponds to -pi/2 phase angle.
     
    Last edited: May 11, 2016
  10. May 11, 2016 #9
    isn't it z^3 = -3iz* ?

    then r^3 = -3r, if i put i into the parentesis of z*
    ==> r^2 = -3r, but then it's imaginary again
     
  11. May 11, 2016 #10

    ehild

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    yes, it is.
    No. The magnitude can not be negative, and i z* ≠ -z. The magnitude of the product of complex numbers is equal to the product of the magnitudes. What is the magnitude of z? What is the magnitude of z*? What is the magnitude of i?
    (Do you know the exponential form of complex numbers?)
     
  12. May 11, 2016 #11
    i think, yeah.

    |z*| = |z|
    |z^3| = |-3iz*| ==> |z|^3 = |-3i| * |z| ==? |z|^2 = |-3i| = 3 ==? |z| = sqrt(3)
     
  13. May 11, 2016 #12

    ehild

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    Correct! And r is the magnitude of z, so r=√3.
    Now determine the phase angles at both sides. The angles add in case of a product. What are the phase angles of z* and -i?
     
  14. May 11, 2016 #13
    -i*z = (cos(3/2pi) + isin(3/2pi))*sqrt(3)(cosx + isin-x). = sqrt(3)(cos(x+3/2pi)) + isin(3/2pi-x ))
     
  15. May 11, 2016 #14

    ehild

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    Not quite.
    Try to use the exponential form. z=re, z*=re-iθ, -i = e-iπ/2. Remember you have the equation z3=-iz* to solve.
     
  16. May 11, 2016 #15
    We haven't learned this one yet, thought you meant somthing else. we have learned de moivre formula and it's properties
     
  17. May 11, 2016 #16

    ehild

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    OK, it is a bit more complicated. But you know that the angles add if you multiply two complex numbers. How can you write -i z*?
    z*=r(cos(-θ)+isin(-θ)) and -i=isin(-π/2) .
     
  18. May 11, 2016 #17

    ehild

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    Almost:)
    -3i z* =3 (cos(3/2pi) + isin(3/2pi))*sqrt(3)(cos(-x) + isin(-x)) = 3sqrt(3)(cos(-x+3/2pi)) + isin(3/2pi-x )).
    So the phase angle of the right-hand side is -x+3pi/2 +2pi*k.
    And what is the phase of the left-hand side?
     
  19. May 11, 2016 #18
    well, -i = 1(cos 3/2 π + isin 3/2 π)
    z* = sqrt(3)( cosx + isin -x )
    -i*z* = sqrt(3)(cos(3/2π+ x) + isin(3/2π -x))
    so the product will be as written above, and we could compare it to left handside which is 9(cos 3x +isin3x)
     
  20. May 11, 2016 #19
    Left hand-side, you mean by z^3 ? that would be 9(cos3x + isin3x)
     
  21. May 11, 2016 #20

    ehild

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    The argument of both the cosine and sine must be the same in Moivre formula. Write z* as
    z* = sqrt(3)( cos(-x) + isin (-x )). Cos(-x) is the same as cos(-x)!
    The left-hand side is (√3)3 (cos 3x +isin3x), the right-hand side is 3sqrt(3)(cos(3/2π- x) + isin(3/2π -x)), and you equated the magnitude of both sides already!
    The phase of the left side is equal to the phase of the right side + 2πk.
     
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