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## Homework Statement

Let f be an analytic function on the disc |z|<1 and satisfies |f(z)|≤M if |z|<1.

Show that [itex] |f(z)| \le M \left| \frac{z-a}{1-a'z} \right| [/itex] when |z|<1

where a' is the complex conjugate of a

## Homework Equations

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

## The Attempt at a Solution

Since |z|<1 and consequently |f(z)|≤1, I rewrite the inequality to be proved as

|f(z)|/M ≤ |[itex]\frac{z-a}{1-a'z}[/itex]|

If M=0, the inequality holds, since it is a trivial case of 0 =< 0. If not, we can use the fraction above. Since [itex]|f(z)| \le M, \; \frac{|f(z)|}{M} \le 1[/itex]

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, [itex] \frac{e^{i\theta} - a}{1-a'e^{i\theta}} = e^{i\theta} \frac{1-ae^{-i\theta}}{1-a'e^{i\theta}} [/itex]

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

Here's where it gets confusing. [itex]|e^{i\theta}|[/itex] 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?