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Quantity of zeros on a ring - Complex Analysis

  1. Nov 24, 2013 #1
    Determine the quantitiy of zeroes of the function:
    [itex]f(z)=z^{4}-8z+10[/itex]
    a) Inside the circle [itex]| z | < 1[/itex]
    b) Inside the ring [itex]1 \leq | z | < 2[/itex]


    a)
    [itex]f(z)=(z^{4}-8z)+10=g(z)+h(z)[/itex]
    As [itex]|h(z)| \geq |g(z)| \forall z : | z | = 1[/itex]
    Then by Rouche's Theorem the number of zeros of the function inside the circle is the same than [itex]h(z)[/itex] (0 zeroes).
    b) My idea was to calculate the number of zeroes in [itex]| z | < 2[/itex] and substract the number of zeroes in [itex]| z | < 1[/itex].
    But I can't find a pair of functions to use Roche's Theorem.
    Any hint?
    I ploted here:
    http://www.wolframalpha.com/input/?i=z^4-8z%2B10
    And I'm starting to suspect that the statement is wrong, the roots are very close to the circle and I guess that this is what make the functions so hard to find.
    But I may be wrong and perhaps it's a easy way to decompose [itex]f(z)[/itex]
     
  2. jcsd
  3. Nov 24, 2013 #2

    mfb

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    Staff: Mentor

    (a) is correct.
    Which statement?

    As all roots appear in pairs, you just have to distinguish between three cases for the numbers of roots inside. It looks like a clever contour integral can help, but I'm not sure how exactly.
     
  4. Nov 24, 2013 #3
    The function, because b) is very hard =/ the modulus of a pair of roots is 1.993.
     
  5. Nov 24, 2013 #4

    mfb

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    WolframAlpha gives solutions that are well separated from 2 in their magnitude.
     
  6. Nov 24, 2013 #5
    Sorry... I forget to take the square root, so it isn't 1.993, its 1.412.
    Thank you, I'll continue trying (I tried 5-6 decompositions until I used wolframalpha and I thought it was way harder than it's).
     
  7. Nov 25, 2013 #6
    My professor said that the exercise was miswritten, it was meant to be easier to use Rouche's theorem than others method.
     
  8. Nov 26, 2013 #7
    Very good then. I hate it when I can't do these. Ask your professor to let us to this one instead (same problem):

    $$z^7-5z^3+12$$
     
    Last edited: Nov 26, 2013
  9. Nov 26, 2013 #8
    Around the same contour this one is easy, because [itex]| 2^{7} | > | 40 | + | 12 | \geq | 5z^{3} + 12 |[/itex] (When [itex]| z | = 2[/itex])
    (Triangular inequality).
    Then you have 7 zeroes here (As [itex] z^{7} [/itex] has 7).
    As you have:
    [itex]| 12 | > | 5 | + | 1 | \geq | 5z^{3} + z^{7} |[/itex] (When [itex]| z | = 1[/itex]) you have 0 zeroes in the circle of radius 1.
    Then all the zeroes are in the ring of outer radius 2 and inner radius 1.
    (If the zeroes are n in the circle of radius 1 and m in the circle of radius 2, in the ring they're m-n).
     
    Last edited: Nov 26, 2013
  10. Nov 26, 2013 #9
    Very good. If the objective of this exercise is to expose and teach you Rouche's Theorem, then going through just this easy one is much better in my opinion then a lesson in frustration with one we can't do. :)
     
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