I'm reading Ahlfors, Complex Analysis, pag. 135....he's generalizing Schwarz' Lemma, which states that if $f$ is analytic in the unit disc with $f(0)=0$ then $|f(z)|\leq |z|$. He says...."still more generally we may replace condition $f(0)=0$ by an arbitrary condition $f(z_0)=w_0$ with $|z_0| So far i understood everything. Then Ahlfors states....explicitly this inequality can be written in the form $$\left|\frac{M(f(z)-w_0)}{M^2-\overline w_0f(z)}\right|\leq\left|\frac{R(z-z_0)}{R^2-\overline z_0 z}\right|$$ I can't understand this last expression! Some help?
general form of Schwarz's Lemma
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complex-analysis
1 Answers
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What you have to show is that the maps $S$ and $T$ given by $$S(z):=\frac{M(z-\omega_0)}{M^2-\bar \omega_0z};\quad T(z):=\frac{R(z-z_0)}{R^2-\bar z_0z}.$$ It's enough to do it for $T$. What we have to show is that $|T(z)|<1$ if $|z|
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0very easy explanation, thank you – 2012-12-09