Schwarz Lemma
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High Quality Content by WIKIPEDIA articles! This is an easy consequence of the Schwarz Pick theorem mentioned above: One just needs to remember that the Cayley transform W(z) = (z i)/(z + i) maps the upper half-plane H conformally onto the unit disc D. Then, the map W o o W 1 is a holomorphic map from D onto D. Using the Schwarz Pick theorem on this map, and finally simplifying the results by using the formula for W, we get the desired result. Also, for all z H, frac{left f'(z)right }{mbox{Im }f(z)} le frac{1}{mbox{Im }(z)}.

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High Quality Content by WIKIPEDIA articles! This is an easy consequence of the Schwarz Pick theorem mentioned above: One just needs to remember that the Cayley transform W(z) = (z i)/(z + i) maps the upper half-plane H conformally onto the unit disc D. Then, the map W o o W 1 is a holomorphic map from D onto D. Using the Schwarz Pick theorem on this map, and finally simplifying the results by using the formula for W, we get the desired result. Also, for all z H, frac{left f'(z)right }{mbox{Im }f(z)} le frac{1}{mbox{Im }(z)}.