# Schwarz's lemma

Suppose that f is analytic in the disc |z|<1, that f(0) = f'(0) = 0 and that |f(z)| ≤ 1 for all z in the disc. Show that |f(z)| ≤ |z^2|, for |z|<1.

Hint: Schwarz's Lemma:

Suppose that f is analytic in the disc |z|<1, that f(0) = 0 and that |f(z)| ≤ 1 for all z in the disc.

Then |f(z)| ≤ |z|, for |z|<1.

Equality can hold for some z ≠ 0 only if f(z) = λz, where λ is a constant of absolute value 1.

https://brainmass.com/math/analytic-geometry/schwarz-s-lemma-181457

#### Solution Summary

This uses Schwarz's Lemma to so details about functions.

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