# general form of Schwarz's Lemma

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 $|z_0|<R$ and $|w_0|<M$. Let $\zeta=T(z)$ be a linear transformation which maps $|z|<R$ onto $|\zeta|<1$ with $z_0$ going to the origin, and let $S(w)$ be a linear transformation with $S(w_0)=0$ which maps $|w|<M$ onto $|S(w)|<1$. It is clear that the composite $S\circ f\circ T^{-1}$ satisfies the original form of Schwarz's Lemma, hence we have $|SfT^{-1}(\zeta)|\leq|\zeta|$ or $|Sf(z)|\leq|T(z)|$. "

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?

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What you have to show is that the maps $S$ and $T$ given by $$S(z):=\frac{M(z-\omega_0)}{M^2-\bar \omega_0z};\quad T(z):=\frac{R(z-\omega_0)}{R^2-\bar \omega_0z}.$$ It's enough to do it for $T$. What we have to show is that $|T(z)|<1$ if $|z|<R$. To see that, note that for $|z|=R$, we have $|T(z)|=1$. We conclude by maximum modulus principle.

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very easy explanation, thank you – Federica Maggioni Dec 9 '12 at 19:12