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Integrating over the unit circle

  1. Dec 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that the function ##f(z)## is analytic and that ##|f(z)| \le 1## for all ##|z| = 1##.

    2. Relevant equations


    3. The attempt at a solution
    I was hoping someone could verify my work.

    Okay, if I understand correctly, ##|f(z)| \le 1## is true for all all complex numbers ##z## on the unit circle. In my book we have a theorem which states that the derivative of ##f(z)## can be computed by calculating an integral; that is,

    ##f'(z_0) = \frac{1}{2 \pi i} \int_C \frac{f(z)}{(s-z_0)^2} ds##

    In this particular case, ##z_0 = 0##, which gives us

    ##f'(0) = \int_C \frac{f(z)}{s^2} ds##

    where we will integrate over the unit-circle. Taking the modulus of both sides,

    ##|f'(0)| = | \int_C \frac{f(z)}{s^2} ds|##

    There is a theorem which states that ##|\int_C f(z) dz| \le \int_C |f(z)| dz##. Using this we get

    ##|f'(0)| \le \int_C |\frac{f(z)}{s^2} | ds \iff##

    ##|f'(0)| \le \int_C \frac{|f(z)|}{|s^2|} ds##

    Because we are integrating over the unit circle, ##|f(z)| \le 1## over the unit circle; substituting this larger quantity in, we get

    ##|f'(0)| \le \int_C \frac{1}{|s^2|} ds##.
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    I am stopping here, because, as I write it, it no longer seems correct.
     
  2. jcsd
  3. Dec 12, 2014 #2

    Orodruin

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    What happens to the differential dz when you take the absolute value inside the integral?
    What happened to your pre factor of 1/(2πi)?
    What is |s^2| on the unit circle?
     
  4. Dec 12, 2014 #3
    hmmm...I am not sure...##|dz| = \sqrt{(dx)^2 +(dy)^2}##?

    Ah! I always seem to neglect that portion.

    ##|s^2| = |s|^2## but ##|s| \le 1## so ##|s^2| \le 1##
     
  5. Dec 12, 2014 #4

    Orodruin

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    Yes, you must take the absolute value also of the differential. I am assuming you are using z = x + iy here.

    No, we are on the curve that is the unit circle, not in the unit disk (which is an area and not a curve).
     
  6. Dec 13, 2014 #5
    Ah, of course. So ##|s^2|## would be ##1## on the unit circle.
     
    Last edited: Dec 13, 2014
  7. Dec 13, 2014 #6

    Orodruin

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    Correct. Putting these pieces together, you should now be essentially done.
     
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