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Integrating over the unit circle

  • Thread starter Bashyboy
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  • #1
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Homework Statement


Suppose that the function ##f(z)## is analytic and that ##|f(z)| \le 1## for all ##|z| = 1##.

Homework Equations




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.
 

Answers and Replies

  • #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?
 
  • #3
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What happens to the differential dz when you take the absolute value inside the integral?
hmmm...I am not sure...##|dz| = \sqrt{(dx)^2 +(dy)^2}##?

What happened to your pre factor of 1/(2πi)?
Ah! I always seem to neglect that portion.

What is |s^2| on the unit circle?
##|s^2| = |s|^2## but ##|s| \le 1## so ##|s^2| \le 1##
 
  • #4
Orodruin
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hmmm...I am not sure...##|dz| = \sqrt{(dx)^2 +(dy)^2}##?
Yes, you must take the absolute value also of the differential. I am assuming you are using z = x + iy here.

##|s^2| = |s|^2## but ##|s| \le 1## so ##|s^2| \le 1##
No, we are on the curve that is the unit circle, not in the unit disk (which is an area and not a curve).
 
  • #5
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No, we are on the curve that is the unit circle, not in the unit disk (which is an area and not a curve).
Ah, of course. So ##|s^2|## would be ##1## on the unit circle.
 
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  • #6
Orodruin
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Ah, of course. So ##|s^2|## would be ##1## on the unit circle.
Correct. Putting these pieces together, you should now be essentially done.
 

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