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Complex polynomial

  1. Dec 4, 2015 #1
    1. The problem statement, all variables and given/known data
    It is known that roots of complex polynomial:

    ##P_n (z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0##

    are the following complex numbers:

    ##\alpha_1, \alpha_2, \cdots, \alpha_n ##

    Find the product:

    ##\prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)##
    2. Relevant equations


    3. The attempt at a solution
    I am pretty sure that i have to use Vieta formulas somehow, and since roots are all complex then it means that there are n/2 pairs of complex-complex conjugate roots. If i multiply all of this i would end up with product of all roots ##\alpha_1* \alpha_2* \cdots* \alpha_n ## and by Vieta formulas i can easily determine that one, but there are more elements of this product after multiplying, what can i do with them?
     
  2. jcsd
  3. Dec 4, 2015 #2
    Ok so it has ##n## roots without counting multiplicity.
    So your polynomial has the form ##P_n(z) = A(z) (z-\alpha_1)...(z-\alpha_n) ##.
    Explain why ##A = 1##
     
  4. Dec 4, 2015 #3

    HallsofIvy

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    This is only true if all coefficients are real numbers but you said "complex polynomial" which means that is not necessarily true.

     
  5. Dec 4, 2015 #4

    Ray Vickson

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    The requirement of having complex-conjugate root pairs is true only if all the coefficients are real, and can fail if some of the coefficients are non-real complex numbers. For example, the equation ##x^2 + (2+i)x - 3 i = 0## has roots ##\alpha_1 \doteq 0.522 +0.788\, i## and ##\alpha_2 \doteq -2.552 - 1.788\, i ##.
     
  6. Dec 4, 2015 #5
    I am not really sure, is it maybe because i have coefficient 1 with zn?
     
  7. Dec 4, 2015 #6
    yes, but if you want to convince yourself, you can justify that ## \text{deg}{(A)} = 0##, so that ##A## is a non-zero constant, and with your argument above, ##A = 1##.

    Ok, so now you have the answer, what is ##P_n(-1)## ?
     
  8. Dec 4, 2015 #7
    Well, it's ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##
    But how this gets me any closer to the solution?
     
  9. Dec 4, 2015 #8
    You must have worked on this problem for too long :-)
    Think about it 10 more minutes

    EDIT: or don't think about it the next 10 minutes :-)

    EDIT (+10 minutes) : What did we just say ? ##P_n(z) = (z-\alpha_1)...(z-\alpha_n)##. So what is ##P_n(-1)##?
     
    Last edited: Dec 4, 2015
  10. Dec 4, 2015 #9
    I am sorry, i really don't know anything than this ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## which is obvious and it still isn't clear to me how this is going to help me solving this.
     
  11. Dec 4, 2015 #10

    Ray Vickson

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    Does this formula (up to a possible sign) express your desired product in terms of the coefficients ##a_0, a_1, \ldots, a_{n-1}##? Did the problem ask you to do more than that?
     
  12. Dec 4, 2015 #11
    I really don't understand what are you trying to say, do you mean that ##a_0 + a_1 + \ldots + a_{n-1} = \prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1) ##?
     
  13. Dec 4, 2015 #12

    Samy_A

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    You know that ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##.

    You also know another way to find ## P_n(-1)##:
    Now put the two together.
     
  14. Dec 4, 2015 #13

    Ray Vickson

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    No, absolutely not. YOU said that ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0 ## (not ##\sum a_j##). My question amounts to asking: why do you want to compute ##P_n(-1)##? What do you gain by doing that? How is that related to the product you started with?
     
  15. Dec 5, 2015 #14
    Let's see: I have ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## and ## P_n(-1) =(-1-\alpha_1) \cdots(-1-\alpha_n)##

    Now, taking all -1's out i have ##P_n(-1) =(-1)^n(\alpha_1+1) \cdots(\alpha_n+1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##

    Multiplying both sides by ##(-1)^{-n}## i have:

    ##(\alpha_1+1) \cdots(\alpha_n+1) = 1 + a_{n-1}(-1)^{-1} + \cdots + a_1(-1)^{1-n} + a_0(-1)^{-n}##

    Now, if this is correct i have another question, my idea was to use Vieta's formulas, i mean i have a given product which i have to find:

    ##\prod=(\alpha_1 + 1)\cdots(\alpha_n+1)## i am sure that there would be ##\alpha_1\alpha_2\cdots\alpha_n## but what is with rest of it , i mean i would have 1 at the end, but in order to use Vieta's formulas i should get something like ##\alpha_1\alpha_2 + \alpha_1\alpha_3 + \cdots + \alpha_1\alpha_n + + \alpha_2\alpha_3 + \alpha_2\alpha_4 +\cdots + \alpha_{n-1}\alpha_n## and then sum of products of three and so on all the way to the product of all of them. Is this correct?
     
  16. Dec 5, 2015 #15

    Samy_A

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    Correct.

    Using Vieta's formulas should also work, as you indeed get all possible combinations of products of roots when you expand ##\prod(\alpha_1 + 1)\cdots(\alpha_n+1)##.
     
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