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Zeros of complex function

  1. Sep 20, 2015 #1

    jjr

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    1. The problem statement, all variables and given/known data
    Let ##D={z : |z| <1}##. How many zeros (counted according to multiplicty) does the function ##f(z)=2z^4-2z^3+2z^2-2z+9## have in ##D##? Prove that you answer is correct.

    2. Relevant equations


    3. The attempt at a solution

    The function has no zeros in ##D##, which can be seen quite easily because ##2z^4-2z^3+2z^2-2z>-9## when ##|z|<1##. I am having some trouble proving this though. A suggestion I have is to use the inequality ##|2z^4-2z^3+2z^2-2z|<8##. I can get as far as the next step ##|z^4-z^3+z^2-z| < 4##, but I am not sure how to go on from this point.

    Any hints or tips would be greatly appreciated!

    Sincerely,
    J
     
  2. jcsd
  3. Sep 20, 2015 #2

    Krylov

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    At this point I would invoke Rouché's theorem.
     
  4. Sep 20, 2015 #3

    mfb

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    Staff: Mentor

    Triangle inequality?
     
  5. Sep 21, 2015 #4

    jjr

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    Thanks! So if I understand correctly:

    ##|z^4-z^3+z^2-z| \leq |z^4| + |z^3| + |z^2| + |z| \leq 4 ## because ##|z^n|<1## if ##|z|<1##, where ## n \in {1,2,3,4} ##. And because ##8<9## it's proven.

    Sincerely,
    J
     
  6. Sep 21, 2015 #5

    Krylov

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    Yes, exactly! (Minor detail: I would write "because ## |z^n| \le 1## for ##z \in \partial D## and ##n = 1,2,3,4##" since in the formulation of the theorem the estimate is actually required to hold only on the boundary ##\partial D## of ##D##.)
     
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