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Complex numbers polynomial divisibility proof

  1. Apr 24, 2014 #1
    I'm not sure whether this should go in this forum or another. feel free to move it if needed
    1. The problem statement, all variables and given/known data
    Suppose that [itex] z_0 \in \mathbb{C}. [/itex] A polynomial [itex] P(z)[/itex] is said to be dvisible by [itex] z-z_0 [/itex] if there is another polynomial [itex] Q(z)[/itex] such that [itex] P(z)=(z-z_0)Q(z). [/itex]
    Show that for every [itex] c \in\mathbb{C} [/itex] and [itex] k \in\mathbb{N} [/itex], the polynomial [itex] c(z^k - z_0^k ) [/itex] is divisible by [itex] z - z_0 [/itex]

    2. Relevant equations

    3. The attempt at a solution

    This is just the part a of the problem.
    I havent tried the other parts yet, because Im not sure if my approach was correct.
    well here it is:

    basically I interpreted the problem as: find [itex] Q(z) [/itex] such that [itex] P(z) = (z-z_0)Q(z)[/itex] and [itex] P(z) = c(z^k - z_0^k ) [/itex]

    then [itex] Q(z) = \Large\frac{P(z)}{z-z_0)} = \Large c\frac{(z^k - z_0^k)}{z-z_0} [/itex]
    then using complex division (AKA multiply by the conjugate):
    [itex] Q(z) = \Large c\frac{z^k(z+z_0) }{z^2 + z_0^2} -\Large c\frac{z_0^k(z+z_0) }{z^2 +z_0^2} [/itex]

    [itex] Q(z) = \Large\frac{cz^{k+1} +cz^kz_0}{z^2 +z_0^2} - \Large\frac{cz_0^kz+cz_0^{k+1}}{z^2+z_0^2} [/itex]

    and then: [itex] Q(z) = \Large\frac{c}{(z^2 +z_0^2)}\normalsize(z^{k+1} + z^kz_0 -z_0^kz - z_0^{k+1}) [/itex] which is a polynomial.

    So: there exists a polynomial satisfying the required conditions and therefore [itex] P(z)[/itex] is divisible by [itex]z-z_0[/itex]


    it looks okay but I'm not 100% confident on this...
     
  2. jcsd
  3. Apr 24, 2014 #2

    HallsofIvy

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    I think you will need to give an argument that this is a polynomial!

     
  4. Apr 24, 2014 #3
    Well...
    now that you mention it I'm not quite sure I have an argument for that...
     
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