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Complex number question

  1. Aug 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi guys, I have no idea what this question wants me to do. Any clarification would be appreciated.

    Let z0, z1, z2, z3 and z4 be the solutions that you obtained. Use the factorization

    z5+1= (z - z0) (z - z1) (z - z2) (z - z3) (z - z4)

    to determine the complex number that is obtained by multiplying together all the solutions of the equation z5= -1.

    3. The attempt at a solution

    These are the solutions I managed to calculate

    z0= cos(pi/5) + i sin(pi/5)

    z1= cos(3pi/5) + i sin(3pi/5)

    z2= cos(pi) + i sin(pi)

    z3= cos(7pi/5) + i sin(7pi/5)

    z4= cos(9pi/5) + i sin(9pi/5)

    Does the question want me to plug these in and multiply them out because there's no chance that's happening or am I missing something?

    Thanks in advance. :smile:
     
  2. jcsd
  3. Aug 30, 2014 #2

    nrqed

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    The question is confusing to me. It sounds like they want you to multiply the solutions to get a complex number, so it does not sounds like they want you to recover the polynomial ##z^5+1 ##. In that sense you won't need that formula.
    :biggrin: :biggrin:

    It looks like you are on the right path. However, do you know about the complex exponential ## E^{i \theta} ##? All your solutions can be written very simply in terms of that exponential and in that case, multiplying all the roots takes one line!
     
    Last edited: Aug 30, 2014
  4. Aug 30, 2014 #3
    Thanks for your reply. :smile:

    I've only briefly come across the complex exponential. I know it expresses a complex number much more concisely but usually the questions I get asked specify if that's what they're looking for. To be honest there's nothing about how to answer the question in any of the textbooks given so I'm not sure how to approach it.
     
  5. Aug 30, 2014 #4

    nrqed

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    You are welcome.

    The key identity is

    [itex] e^{i \theta} = \cos (\theta) + i \sin(\theta) [/itex]

    Now you see how each of your roots can be expressed as a single exponential.
    The nice thing now is that multiplying exponentials is trivial!
     
  6. Aug 30, 2014 #5

    Mark44

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    nrqed, maybe you typed '-' when you meant '='. The roots of the polynomial equation z5 + 1 = 0 are the values for z0, z1, z2, z3, and z4 in the factored form.
     
  7. Aug 30, 2014 #6

    nrqed

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    Thanks, I actually meant ##z^5+1 ##. My mistake.
     
  8. Aug 30, 2014 #7

    vela

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    I think the idea is to infer what the product ##z_0 z_1 z_2 z_3 z_4## equals by comparing the two sides of the equation. It's a conceptual problem, not a grind-it-out problem.
     
  9. Aug 31, 2014 #8

    nrqed

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    Agreed, that works and is more elegant (and probably what they wanted)
     
  10. Aug 31, 2014 #9
    Thanks guys.

    I'm still not entirely sure what I need to do in respect to comparing the two sides to see what the product equals. Is there any worked examples of similar problems I can look at?

    Thanks.
     
  11. Aug 31, 2014 #10

    ehild

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    You can factorize a polynomial P(x) if you know the roots of the equation P(x)=0.
    In case of a quadratic equation,

    x2+bx+c=(x-x1)(x-x2)=0.

    If you can factorize P(x) this way, x1 and x2 are the roots. Expanding the product you get x2-(x1+x2)x +x1x2.
    The two sides are equivalent if the coefficients of the equal powers of x are the same at both sides.
    For x2, it is true, both are 1.
    The coefficient of x is b on the left side and it is -(x1+x2) on the right side, so b=-(x1+x2).
    The constant term is c on the left side and (-x1)(-x2) on the right side, so c=(-x1)(-x2).

    In the same way, if you have a complex polynomial of fifth order you can factorize it and the constant term must be the same of both sides.

    See Vieta's Formulas http://mathworld.wolfram.com/VietasFormulas.html

    Your polynomial is z5+1 and the roots of the equation z5+1=0 are z1,z2,z3,z4,z5.

    z5+1=(z-z1)(z-z2)(z-z3)(z-z4)(z-z5).
    When expanding the right side, all terms contain some power of z except the last, the product of the constant terms - what is that product?
    It must be the same as the constant term at the left side. What do you get?


    ehild
     
  12. Aug 31, 2014 #11
    Thanks for replying. :smile:

    I've not come across this before so sorry if I've completely missed the point. So is this what the question is looking?

    (z+1)*(z4-z3+z2-z+1)
     
  13. Aug 31, 2014 #12

    ehild

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    It is the left side. I meant the other side. If you multiplied all the 5 factors, what would be the term that contains only z1,z2, z3, z4, z5 and no z?

    ehild
     
  14. Aug 31, 2014 #13
    I don't think so. It just wants you to multiply all the roots you found. (Don't "plug them in", just take the 5 numbers and multiply them.) nrqed recommended representing the roots you found in ##e^{i \theta}## form. That will make it easier, but the result should not be surprising either.
     
  15. Aug 31, 2014 #14

    ehild

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    It is the left side. I meant the other side.

    If you expand a product (x+a)(x+b)(x+c)(x+d)(x+e) then you get a single term with x5

    terms with x4: (-a-b-c-d-e)x4

    terms with x3 : x3(ab+ac+ad+ae+.... )

    terms with x2: x2(abc+abd+abe+bcd+.....)

    terms with x1: x(abcd+abce+bcde+....)

    at the end you have one constant term: abcde

    In the problem, a, b, c, d, e are -z1, -z2, -z3, -z4, -z5.
    The constant term is : ?

    The constants are the same on both sides: The left side is z5+1, the constant is 1. That is equal to the constant on the right side which is ??

    ehild
     
  16. Aug 31, 2014 #15

    vela

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    Multiply out the righthand side symbolically. What do you get? You should see a pattern to the coefficients, and you should see why this pattern emerges.
     
  17. Aug 31, 2014 #16
    So is it -1? :confused:
     
  18. Aug 31, 2014 #17
    Also, I think you are going to be getting two different responses to how to handle this.

    I am answering your issue based on the fact that you have already provided a certain amount of work. You found the roots in step 3. Hence, all you have to do is multiply them together.

    Others may answer your question in the more general sense. The problem statement doesn't require you to find the exact roots like you did in step 3. You could just examine the algebra provided in step a to find the answer.
     
    Last edited: Aug 31, 2014
  19. Aug 31, 2014 #18

    PeroK

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    Does this help?

    ##z^5 + 1 = (z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4) = z^5 -(z_0+z_1+z_2+z_3+z_4)z^4+?z^3+?z^2+?z+?##

    It certainly tells you something about the sum of the roots. And perhaps you can see something about the product of the roots as well?
     
  20. Aug 31, 2014 #19

    ehild

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    What is -1? :smile:If you mean the solution of the problem, yes, it is: The constant term on the right side is (-z1)*(-z2)*(-z3)*(-z4)*(-z5)= -z1z2z3z4z5=1, so the product of the roots is -1.

    ehild
     
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