# Integrating over the unit circle

• Bashyboy
In summary, we are given that the function ##f(z)## is analytic and satisfies ##|f(z)| \le 1## for all complex numbers ##z## on the unit circle. We want to compute the derivative of ##f(z)## at ##z=0## using the formula ##f'(0) = \frac{1}{2 \pi i} \int_C \frac{f(z)}{(s-z_0)^2} ds##. By taking the modulus of both sides, using the theorem ##|\int_C f(z) dz| \le \int_C |f(z)| dz##, and noting that ##|s^2| = 1## on the unit circle, we can

## Homework Statement

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

## 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.

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?

Orodruin said:
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}##?

Orodruin said:
What happened to your pre factor of 1/(2πi)?

Ah! I always seem to neglect that portion.

Orodruin said:
What is |s^2| on the unit circle?

##|s^2| = |s|^2## but ##|s| \le 1## so ##|s^2| \le 1##

Bashyboy said:
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.

Bashyboy said:
##|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).

Orodruin said:
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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Bashyboy said:
Ah, of course. So ##|s^2|## would be ##1## on the unit circle.
Correct. Putting these pieces together, you should now be essentially done.