Integrating over the unit circle

[Bashyboy]

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.

 

[Orodruin]

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?

 

[Bashyboy]

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 $|s^2| \le 1$

 

[Orodruin]

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).

 

[Bashyboy]

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.

 

[Orodruin]

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

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