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!

Homework Help: Complex analysis question (Roche theorem)

  1. Mar 26, 2008 #1


    User Avatar
    Gold Member

    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.
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook