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Complex analysis question (Roche theorem)

  1. Mar 26, 2008 #1
    Let R be domain which contains the closed circle:
    |z|<=1, Let f be analytic function s.t f(0)=1, |f(z)|>3/2 in |z|=1, show that in |z|<1 f has at least 1 root, and and one fixed point, i.e s.t that f(z0)=z0.


    now here what I did, let's define g(z)=f(z)-z, and we first need to show that the number of roots of f(z) is at least as z (i.e 1), by roche theorem, if |f(z)|<|z-f(z)| for every |z|<1 then the number of roots of f(z) are the same as those of (z-f(z))+f(z), now the pesky task to show this.
    |z-f(z)|<=|z|+|f(z)|<1+|f(z)|
    here is where I'm stuck, don't know how to follow.
     
  2. jcsd
  3. May 29, 2008 #2
    If all you want to do is show that |z-f(z)|<=|z|+|f(z)|<1+|f(z)| is true inside the unit circle then:

    |z-f(x)| <= |z|+|-f(z)| = |z|+|f(z)| by the triangle inequality

    since we are inside the circle, |z| < 1. add |f(z)| to both sides, it is positive so the inequality remains.
    |z| + |f(z)| < 1 + |f(z)|

    On the unit circle, |z|=1 and |f(z)| > 3/2 so there are definitely no zeroes there.
     
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