Word Count : 141

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Let (\phi(z)=\dfrac{2z-1}{2-z}). For (z) on the circle (|z|=1), consider [ |\phi(z)|^2=\frac{|2z-1|^2}{|2-z|^2}. ] Now [ |2z-1|^2=(2z-1)(2\bar z-1)=4z\bar z-2z-2\bar z+1, ] and [ |2-z|^2=(2-z)(2-\bar z)=4-2z-2\bar z+z\bar z. ] Since (|z|=1), _________ _____ _____ _______ _______ ________ __________ ________ _______ _________ ___ ___.
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