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Complex Roots

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve z^5 + 16 conjugate(z) = 0 for z element of C.

    z^5 + 16z' = 0

    http://puu.sh/2EBqC.png [Broken]


    2. Relevant equations



    3. The attempt at a solution

    My first thought was to use z = a+bi and z' = a-bi
    So:
    (a+bi)5 + 16*(a-bi) = 0 + 0i And then expand and simplify to the real and non real parts. http://puu.sh/2EBwU.png [Broken]
    But it seems WAAY too complicated for a question worth only 4 marks.
    I would appreciate if someone could post their thoughts.
    Thanks!


    According to WolframAlpha: http://puu.sh/2EBKH.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 21, 2013 #2

    tiny-tim

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    hi planauts! :smile:

    (try using the X2 button just above the Reply box :wink:)
    try multipying by z :wink:
     
  4. Apr 21, 2013 #3

    Dick

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    I think you'll have better luck with polar form than with rectangular. Try it.
     
    Last edited by a moderator: May 6, 2017
  5. Apr 21, 2013 #4

    BruceW

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    I agree with these guys. Also, there are more solutions than wolfram alpha said. the button "more roots" gives the rest. (just in case you were confused that your roots were not the same as the ones you saw in wolfram alpha).
     
  6. Apr 21, 2013 #5

    [itex]z^6 + 16|z|^2 = 0[/itex]
    [itex](z^3+4|z|i)(z^3-4|z|i) = 0[/itex]
    Factor further using sum and differences of cubes?

    I'm not sure what you mean,

    [itex]r^5e^{5i\theta}+16re^{-i\theta}=0[/itex]
    [itex]r(r^4e^{5i\theta}+16e^{-i\theta})=0[/itex] r = 0
     
  7. Apr 21, 2013 #6

    Dick

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    That's exactly what I mean. Now multiply both sides by ##e^{i\theta}## and start thinking about what ##\theta## and r values might give you a solution.
     
  8. Apr 21, 2013 #7
    I end up with.
    [itex]r^4e^{6i\theta} = -16[/itex]

    I know theta = -pi/2 and r = 2 would give a solution...

    How would I manipulate this equation so I can use the De Moivre theorem?

    Should I let 4w = 6theta

    So
    [itex]r^4*e^{4wi} = -16[/itex]

    Let [itex]p^4 = r*e^{wi} = -16[/itex]
    And then solve for p and change the w angles to theta? I know that r has to be definitely 2 though.

    Am I over complicating things again?
     
  9. Apr 21, 2013 #8

    Dick

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    Maybe. Yes, r must be 2. So ##e^{6i\theta}=(-1)##. That means ##cos(6\theta)=(-1)## and ##sin(6\theta)=0##, right? That's deMoivre. I don't think it should be hard from there.
     
  10. Apr 22, 2013 #9

    tiny-tim

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    hi planauts! :smile:
    no, it's easier now to put z = |z|e instead,

    giving you |z|4e6iθ = -24

    (same as you got anyway)
     
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