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Homework Help: Show that the four of the five roots of

  1. Feb 3, 2012 #1
    Show that the four of the five roots of ....

    1. The problem statement, all variables and given/known data

    Show that four of the five roots of z5 + 15z + 1 = 0 belong to the annulus {z: 3/2 < |z| < 2}.

    2. Relevant equations

    Argument Principle (presumably)

    3. The attempt at a solution

    Since f(z) = z5 + 15z + 1 is entire, it has no poles. Thus, if N1 is the number of zeros inside C2(0), N2 is the number of zeroes inside C3/2(0), and N is the number of zeroes inside the annulus {z: 3/2 < |z| < 2},

    N = N2 - N1 = 1/(2πi) [ ∫C2(0)f'(z)/f(z) dz - ∫C3/2(0)f'(z)/f(z) dz ]
    Right? Or is there an easier way?

    EDIT: Ah, shoot! That assumes that |z|=3/2 and |z|=2 have no zeroes.
    EDIT 2: And also that no zeroes are that |z|>2.

    What's the strategy here?
     
  2. jcsd
  3. Feb 4, 2012 #2

    morphism

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    Re: Show that the four of the five roots of ....

    It's better to use Rouche's theorem here.
     
  4. Feb 4, 2012 #3
    Re: Show that the four of the five roots of ....

    Right, and with it I can show that 5 roots are in the range |z| < 2 and 1 root is in the range |z| < 3/2. Then I just need to show that no roots are on the circles |z| = 2 or |z| = 3/2. Eh?
     
  5. Feb 4, 2012 #4

    morphism

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    Re: Show that the four of the five roots of ....

    So you know that there are 4 roots in 3/2<=|z|<2 (you don't need to worry about |z|=2 - think about what Rouche's theorem says to see why).

    To exclude the possibility of there being a root on the circle |z|=3/2, you can repeat the same Rouche argument with 3/2+epsilon in place of 3/2 to show that there's one root in |z|<3/2+epsilon (for sufficiently small epsilon).
     
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