1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook