How to show that $\dfrac{z-a}{1-\bar{a}z}$ maps $\mathbb{T}$ into $\mathbb{T}$, where $\mathbb{T}=\{z \in \mathbb{C}: |z|=1\}$? How it is possible that just three points is enough to show that?
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1Did you mean $\dfrac{z-a}{1-\overline{a}z}$? – Daniel Fischer Sep 04 '13 at 16:03
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1You should mention what is $\mathbb{T} $. – smiley06 Sep 04 '13 at 16:03
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$,\Bbb T,$ can be lots of things, even if we focus on linear fractional transformations, complex analysis or whatever. To expect all know what you mean is a long shot. – DonAntonio Sep 04 '13 at 16:06
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1If you don't like "three point method" then show that $\left|\dfrac{z-a}{1-\overline{a}z}\right|=1$ for $|z|=1$. – njguliyev Sep 04 '13 at 16:09
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1My guess is that (1) Daniel Fischer was right: $a*$ should be $a^\ast$ or $\overline a$, the complex conjugate, and (2) $\mathbb T$ is the unit circle ${z\in\mathbb C : |z|=1}$. ${}\qquad{}$ – Michael Hardy Sep 04 '13 at 16:21
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I meant $a^*=\bar{a}$. – Sep 04 '13 at 16:57
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Lemma. If $L$ is bilinear maps for which one are points $0$, $\infty$ and $1$ fixed, then $L$ is identic map.
Proof. From $L(\infty)=\infty$ we conclude that $L$ is entire linear function, $$L(z)=Az+B.$$ But, from $L(0)=0$ we get $B=0$, and from $L(1)=1$ that $A=1$. So, $L(z)=z$. $\blacksquare$
Theorem. For every three different points $z_1,z_2,z_3 \in \mathbb{C}$ and three different points $w_1,w_2,w_3$ there exist unique bilinear map $L$ such that $L(z_k)=w_k,\, k=1,2,3.$
Proof. Construct bilinear mapping $L_1$ and $L_2$ which maps $z_1,z_2,z_3$ and $w_1,w_2,w_3$, respectively, in points $0,\infty$ and $1$.
$$L_1(z)=\frac{z-z_1}{z-z_2} \cdot \frac{z_3-z_2}{z_3-z_1}, \quad L_2(w)=\frac{w-w_1}{w-w_2}\cdot \frac{w_3-w_2}{w_3-w_1}.$$
Our map is $L=L_2^{-1} \circ L_1$, that is $w=L_2^{-1} \circ L_1(z).$
So, $L_2(w)=L_1(z)$, that is $$\frac{z-z_1}{z-z_2} \cdot \frac{z_3-z_2}{z_3-z_1}=\frac{w-w_1}{w-w_2}\cdot \frac{w_3-w_2}{w_3-w_1}.$$
Let prove unique of this maps. Let $A$ be bilinear map such that $A(z_k)=w_k \, (k=1,2,3)$. As points $0,\infty$ and $1$ are fixed points of bilinear mapping $B=L_2 \circ A \circ L_1^{-1}$, then Lemma is giving us that $B=E,$ where $E$ is identic map. From that, using laws of group, we get $A=L_2^{-1} \circ L_1$, that is $A=L$. $\blacksquare$
For first question see here.
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Could you please write how is it that for example $\phi_a(i) \rightarrow w \in \mathbb{D}$ $w=?$. What is this $w$ if $a\in \mathbb{D}$? Does it belong to $\mathbb{D}, \forall a\in \mathbb{D}$? – Sep 09 '13 at 15:54