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Complex analysis

  1. Mar 8, 2012 #1
    Two questions:
    1)Quote comes from a textbook:

    Each non-constant function analythic function with f(0)=0 is,in a small nbhd of 0, the composition of a conformal map with the nth-power map...The proof is given and I think I am comfortable with it..

    My question is a lot simpler (I think): Can we say the same for a function such that f(a)=b?
    I am kindly asking someone to explain why one loses no generality if we
    assume a=b=f(a)=0.

    2) If an analytic function is not zero in a nbhd, what can I say about the derivative there? i.e. what restriction does this impose on f' (f prime)

    Around a circle that lies in the nbhd above we should have
    # Zeros=Integral (f'/f)=0

    Should this not say f'=0 on that circle?
    Thank you
     
  2. jcsd
  3. Mar 8, 2012 #2

    morphism

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    1) If f(a)=b, then f(z) looks like b+(z-a)^n near a.

    2) Could you explain why you think f'=0 on the circle?
     
  4. Mar 8, 2012 #3
    I was thinking: if f' were to be never zero, then how can Integral[f'/f] around a circle be zero.
    The integral should be zero as f has no zeros inside the circle...Is what I am thinking.

    Than you for your time again
     
  5. Mar 8, 2012 #4

    morphism

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    The contour integral of a nonzero function can very well be zero. Take for instance f(z)=z-2 and integrate f'/f=1/(z-2) on the unit circle.
     
  6. Mar 8, 2012 #5
    Somewhat embarrassing...

    Thank you very much for the help
     
  7. Mar 13, 2012 #6

    Bacle2

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    Actually, if f is analytic and 1-1, then f' is not zero; if f'(z) is 0 , then factor:

    f(z)-zo=z^k(g(z)) (i.e., use Taylor series, where k is the index of the zero of f ),

    and g(z) is analytic and non-zero in a 'hood U of zo (the zeros of an analytic non-constant

    function are isolated) . The idea is that these conditions on f allow you to define a local k-th root function

    in the ball, and k-th roots are k-to-1, and so in particular not 1-1.


    Under these conditions , take a

    small ball B(z,r) around f(z). This allows you to define a log locally, so that you can

    then define a k-th root of g(z) (technically, the ball is simply-connected, and does not

    wind around 0). Then you can define a k-th root using the log, and k-th roots are not

    1-1; they are actually 1-1. Maybe needs work, but it is late.
     
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