Let f be analytic inside and on the unit circle. Suppose that 0<|f(z)|<1 if |z| = 1.
Show that f has exactly one fixed point inside the unit circle. ( note : a fixed point is a point Zo such that f(Zo) = Zo).
For |z| = 1 we have
|-z| = 1 > |f(z)|
So by Rouche's ...
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