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Complex roots of a polynomial

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Let a0 > a1 > a2 > ... > an > 0 be coefficients of a polynomial P(z) = a0 + a1z + a2z2 + ... + anzn. Let z0 be complex number such that P(z0) = 0. Show that |z0| > 1.


    2. Relevant equations
    Triangle inequality? Not sure if that's enough.


    3. The attempt at a solution
    I started with asumption that P(z0) = 0 with some |z0| < 1. Then I made a pair of equations:

    z0P(z0) = 0
    P(z0) = 0

    Subtraction gives a new equation

    z0P(z0) - P(z0) = 0

    I rearranged the terms

    -a0 + (a0 - a1)z0 + ... + (an-1 - an)z0n + anz0n+1 = 0

    and took a modulus of it. Then by using triangle inequality (and the asumption that |z0| < 1) I got a contradiction 0 > 0. So the modulus of z0 must be greater or equal to 1.

    For the second part of the task, I also tried a similar approach (assumed that |z0| = 1 and tried to make it lead to a contradiction) but couldn't get anything out of it. In the first part I needed the asumption that |z0| < 1 to get the contradiction 0 < 0. Now the same approach leads to inequality 0 [tex]\leq[/tex] 0 which is true.

    Any tips? :smile:
     
  2. jcsd
  3. Feb 7, 2009 #2

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    Not sure if it will work, but I would try estimating it from below. That is, use the triangle inequality [itex]|a|-|b|\leq |a-b|[/itex].
     
    Last edited: Feb 7, 2009
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