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Complex Number, again

  1. Oct 11, 2004 #1

    JasonRox

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    Complex Number, again... :(

    This I'll give you the entire question and answer.

    10. Solve:

    [tex]z^4-2z^2+4=0[/tex]

    That's all I got.

    Answer:

    [tex]+-1/2(\sqrt{6}+-\sqrt{2i}) (four combinations of signs).[/tex]

    That is all.

    I tried factoring, but I can't come up with anything. I also tried bisecting, but that is useless.

    Knowing that the solution has complex numbers, I have no clue where they got this from.
     
  2. jcsd
  3. Oct 11, 2004 #2

    Dr Transport

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    solve for [tex] z^{2} [/tex] then take the square root of those........
     
  4. Oct 11, 2004 #3

    JasonRox

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    Ok then.

    [tex]z^2=(\frac{z^2}{\sqrt{2}}-2i)(\frac{z^2}{\sqrt{2}}+2i)[/tex]

    The square root of these?
     
  5. Oct 11, 2004 #4
    my results are
    [tex]z=+-\sqrt{1+i\sqrt{3}}[/tex]
    [tex]z=+-\sqrt{1-i\sqrt{3}}[/tex]

    I dont know whether they are correct or not..
     
  6. Oct 11, 2004 #5

    HallsofIvy

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    let x= z2 then the equation z4- 2z2+ 4= 0 becomes x2- 2x+ 4= 0. That is the same as x2- 2x= -4 or
    x2- 2x+ 1= (x- 1)2= -3. From that x= [itex]1+- \sqrt{3}i[/itex].
    Since x= z2, [itex]z^2= 1+- i\sqrt{3}[/itex] so [itex]x= +- \sqrt{1+- i\sqrt{3}}.
     
  7. Oct 11, 2004 #6

    JasonRox

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    I can see that, but where would their solution come from?

    Using the quadratic formula yield the same result as yours, too.
     
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