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Application of Rouche's Theorem HELP

  1. Oct 18, 2009 #1
    Question: Find the number of zeroes of the equation [tex]$ z^7 - 2z^5 + 6z^3 - z+1 = 0 $[/tex], in the unit disk.


    We have Rouche's theorem which says that if f(z) and g(z) are two functions analytic in a neighborhood of the closed unit disk and if |f(z) - g(z)| < |f(z)| for all z on the boundary of the disk, then f and g have the same number of zeroes in the disk.

    Now there are easier problems which I can do just fine. But here is what I try. Now first off I know from here-say and mathematica that there are three roots inside the unit disk. So naturally I could use g(z) is either "6z^3" or "-2z^5 + 6z^3" since both have three zeroes (counting multiplicity) INSIDE the disk. So here is all i can see:

    using the identity [tex]$\right|a+b\right| \le \left|a\right|-\left|b\right|$ [/tex]over and over again I get:
    |f(z)| >= 1 immediately not very helpful....
    from below I can see, using [tex]$g(z) = -2z^5 + 6z^3$[/tex],
    [tex]$\left|f(z)-g(z)\right| = \left|z^7 - z + 1\right| < 3$ [/tex] where i deliberately use strict < if you think about the geometry of this triangle inequality application....

    ok so I cant get my |f(z)-g(z)| to be smaller than |f(z)|.... any help?
     
  2. jcsd
  3. Oct 19, 2009 #2
    Try using the alternate (equivalent formulation) of Rouche's theorem
    [tex]|f(z) - g(z)| < |f(z)| + |g(z)|[/tex]

    using [tex]f(z) = z^7 - 2z^5 + 6z^3 - z + 1[/tex] and
    [tex]g(z) = 6z^3[/tex]
    .

    Because then
    [tex]|f(z)-g(z)| = |z^7 - 2z^5 - z + 1|[/tex]
    [tex]|f(z)-g(z)| \leq 5 < 6 \leq |6z^3|[/tex]
    [tex]|f(z)-g(z)| < |z^7 - 2z^5 + 6z^3 - z + 1| + |6z^3| = |f(z)| + |g(z)|[/tex]
    .

    There is a very similar example to your problem in "Handbook of complex variables" by Krantz on page 74.
     
  4. Oct 19, 2009 #3
    Ahh thank you so much. I see that version of the theorem in Ahlfors. Although, I don't see why I would ever use the version of the theorem I had originally quoted since clearly having that extra "[tex]$ + \left| g(z) \right|$ [/tex]" makes these inequalities a heck of a lot easier.
     
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