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Solve a polynomial function

  1. Jan 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Given the polynomial function ##x^4+x^3+2x^2+4=0## solve it if you know that it has at least one complex zero whose real part equals the complex part.
    2. Relevant equations
    3. The attempt at a solution

    My guess is that if this function has one complex zero it must have a conjugate complex zero as well so we need to find those and two more. Im not sure how to start except realising that ##P(b-bi)=P(b+bi)=0## but solving the equation for that seems a lot of work and im sure theres an easir way. Should i use Viet's formulas or some other way to approach this?
     
  2. jcsd
  3. Jan 11, 2017 #2

    Mark44

    Staff: Mentor

    I haven't worked this problem, but here are my thoughts. If x = b + bi is a solution (implying that x = b - bi is also a solution), what will be the effect of the x3 term? Will it be possible for ##x^4 + x^3 + 2x^2## to be equal to the pure real number -4?
     
  4. Jan 11, 2017 #3

    epenguin

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    I was going to say State what you understand by 'Real part equals the complex part' which, I don't know, sounded slightly loose language to me, but what I understood is what doktorwho has said. It is telling you something special about one quadratic factor of the polynomial.

    And there is also something else a bit special about the coefficients, and hence the roots. So lay that out and you have a few things you can may be put together (I also haven't worked the problem).
     
  5. Jan 11, 2017 #4
    It seems that the other two zeros are complex as well. So we have some ##a+zi## and ##a-zi## as zeros.
    ##P(x)=(a+zi)(a-zi)(b+bi)(b-bi)##
    ##P(x)=2b^2 (a^2 + z^2)##
    Doesnt seem right..
    I can also write it in this form but doesnt seem to give any ideas..
    ##(x^2 + x + 2) x^2 + 4 = 0##
     
  6. Jan 11, 2017 #5

    Mark44

    Staff: Mentor

    You didn't follow up on the question I asked, which is, in essence, "Is it possible for b + bi to be a solution?"
    In your work above, you are tacitly assuming that both b + bi and b - bi are solutions.
    Factoring part of one side of an equation is no help.
     
  7. Jan 12, 2017 #6
    It cant be equal to -4. Maybe there's some other methid im not noticing.. Well if one complex zero has the two parts equal shouldnt (##b+bi##) be the right way to state that?
     
  8. Jan 12, 2017 #7

    Ray Vickson

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    I think the problem is wrong: it is impossible to have a root of the form ##x = a + i a## for real ##a##. It is true that one of the roots has a real part that is near the imaginary part, but they are not equal exactly.

    You should be able to show this explicitly: expand out ##P(a + ia)## to find its real and imaginary parts. If you equate both of those parts to zero you will see that there is no solution for ##a##.
     
  9. Jan 12, 2017 #8

    haruspex

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    Are you sure it isn't ##x^4+x^3+2x^2+2x+4=0##?
     
  10. Jan 12, 2017 #9

    Ray Vickson

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    That does work.
     
  11. Jan 12, 2017 #10

    ehild

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    Wolframalpha says that there are no roots with equal real and imaginary parts.
     
  12. Jan 12, 2017 #11

    Mark44

    Staff: Mentor

    Assuming that the equation above is the correct equation (and not as surmised in post #8), the part "if you know that it has at least one complex zero whose real part equals the complex part." is clearly false, so no work is required.
     
  13. Jan 13, 2017 #12
    I just got a response saying that it was a typo, ##2x## term should be included and then ##-1-i## and ##-1+i## are solutions and the rest can be gotten by dividing the polynomial by ##(x-(-1-i))*(x-(-1+1))##.
     
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