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Complex Variable Problem

  1. Jan 11, 2010 #1
    If f is entire, and 1<=[tex]\left|f\right|[/tex] <=2 for all [tex]\left|z\right|[/tex] =1, and there is a z0 with [tex]\left|z0\right|[/tex] <1 and f(z0)=z0, then prove or disprove that there exist a z1 with [tex]\left|z1\right|[/tex]<1 such that f(z1)=0.
  2. jcsd
  3. Jan 11, 2010 #2
    I am thinking of using Rouche's Theorem, but find it not that useful.
  4. Jan 12, 2010 #3
    a bounded entire function is a constant
  5. Jan 12, 2010 #4
    This is only true if it was bounded on the entire complex-plane, here f is not even bounded in a domain, but on a line.
  6. Jan 12, 2010 #5
    right I misread your problem
  7. Jan 12, 2010 #6
    I think your idea of using Rouche' Theorem is right.

    If z0 = 0 there is nothing to prove.
    If z0 is not zero then

    let g(z) = f(z) - z0. |f - g| = |z0| < 1 <= min(f(z) on the unit circle.

    So f and g have the same number of roots inside the unit circle ( f has no roots on the unit circle.)

    I am not sure why you need f to be entire.
  8. Jan 13, 2010 #7
    That's right, thanks!
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