# Integrating over the unit circle

1. Dec 12, 2014

### Bashyboy

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. Dec 12, 2014

### Orodruin

Staff Emeritus
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. Dec 12, 2014

### Bashyboy

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 $|s^2| \le 1$

4. Dec 12, 2014

### Orodruin

Staff Emeritus
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).

5. Dec 13, 2014

### Bashyboy

Ah, of course. So $|s^2|$ would be $1$ on the unit circle.

Last edited: Dec 13, 2014
6. Dec 13, 2014

### Orodruin

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