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Schwarz's lemma

Suppose that f is analytic in the disc |z|<1, that f(0) = f'(0) = 0 and that |f(z)| &#8804; 1 for all z in the disc. Show that |f(z)| &#8804; |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)| &#8804; 1 for all z in the disc.
Then |f(z)| &#8804; |z|, for |z|<1.
Equality can hold for some z &#8800; 0 only if f(z) = &#955;z, where &#955; is a constant of absolute value 1.

Solution Summary

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

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