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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.