# Proving an inequality using maximum modulus

1. Jul 25, 2013

### stgermaine

1. The problem statement, all variables and given/known data
Let f be an analytic function on the disc |z|<1 and satisfies |f(z)|≤M if |z|<1.
Show that $|f(z)| \le M \left| \frac{z-a}{1-a'z} \right|$ when |z|<1

where a' is the complex conjugate of a

2. Relevant equations
This section uses maximum modulus principle, but I really don't understand it.

3. The attempt at a solution
Since |z|<1 and consequently |f(z)|≤1, I rewrite the inequality to be proved as
|f(z)|/M ≤ |$\frac{z-a}{1-a'z}$|
If M=0, the inequality holds, since it is a trivial case of 0 =< 0. If not, we can use the fraction above. Since $|f(z)| \le M, \; \frac{|f(z)|}{M} \le 1$

Then, I write z as e^iθ. Technically I should write e^(iθ) but for the sake of convenience, e^iθ means just that, not e^i * θ.

Now, $\frac{e^{i\theta} - a}{1-a'e^{i\theta}} = e^{i\theta} \frac{1-ae^{-i\theta}}{1-a'e^{i\theta}}$

Note that the numerator is the complex conjugate of the denominator. Then, we can reduce |f(z)| as $|e^{i\theta}|$

Here's where it gets confusing. $|e^{i\theta}|$ was stipulated to be less than 1 in the problem, but I wish to show that it is greater than equal to one.

What should I do?

2. Jul 25, 2013

### dirk_mec1

Ok, here is how I did it:

1) Multiply with the complex conjugate of the denominator;
2) Use the fact that |z|<1 to get of an 'a';
3) Use |z|<1 again to get rid of the fraction.