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Find all linear fractional transformation of the complex plane that maps the open unit disk {$z:|z|<1$} to the open unit disk {$w:|w|<1$}

What I find is the map $w=z$

Can you help me find another transformation?

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  • $\begingroup$ Daniel//yeah I mistook it. so can you help me find another linear fractional transformation? $\endgroup$ – user114952 Jan 28 '14 at 21:25
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Let $D=\{|z|<1\}$ be the open unit disk. We will prove:

Theorem: All biholomorphic maps $f:D\rightarrow D$ have the form $$f(z)=\alpha \frac{z-a}{1-\overline{a}z}$$ for some $\alpha\in\partial D$ and $a\in D$.

Proof: The first observation is that

$$g_a(z)=\frac{z-a}{1-\overline{a}z}$$ where $a\in D$, maps $D$ biholomorphically onto $D$. Indeed, $g_a$ is holomorphic in $D$ and for $|z|=1$ we have

$$|g_a(z)|=\frac{|z-a|}{|1-\overline{a}z|}=\frac{|\overline{z}-\overline{a}|}{|1-\overline{a}z|}=\frac{|\overbrace{z\overline{z}}^{=1}-\overline{a}z|}{|1-\overline{a}z|}=1$$

Therefore $|g_a(z)|=1$ for $|z|=1$. By the maximum principle we conclude $|g_a(z)|<1$ for $|z|<1$. Thus $g_a$ maps $D$ on $D$. It is bijective since the inverse map is given by

$$g_a^{-1}(z)=\frac{z+a}{1+\overline{a}z}$$

The second observation is the following surprising lemma:

Lemma: Let $f:D\rightarrow D$ be biholomorphic with $f(0)=0$. Then $f(z)=\alpha z$ for some $\alpha\in\partial D$.

Proof: Setting $g(z)=f(z)/z$ we obtain again a holomorphic function on $D$ (it is holomorphic at $0$, since $f(0)=0$). For $|z|=r<1$ we have $|g(z)|=|f(z)|\le 1/r$.

By the maximum principle (this is the key point!) it follows that $|g(z)|\le 1/r$ for all $|z|\le r$ (not just $|z|=r$). Letting $r\rightarrow 1-$ we see that $|g(z)|\le 1$ for all $z\in D$. Therefore we have shown

$$|f(z)|\le |z|\tag{1}$$ for all $z\in D$.

Since $f$ is biholomorphic, the same argument applies to $f^{-1}$, showing that also

$$|f^{-1}(z)|\le |z|$$

for all $z\in D$. But that implies $|z|=|f^{-1}(f(z))|\le |f(z)|$. Combined with $(1)$ this gives $|f(z)|=|z|$, i.e. $|g(z)|=1$. But that means that $g$ attains its maximum in the open unit disk, so by the maximum principle it must be constant: $g(z)=\alpha$ for some $\alpha$ with $|\alpha|=1$. With other words

$$f(z)=\alpha z$$

for $z\in D$. $\square$

Now we can put both together to finish the proof. Let $f:D\rightarrow D$ be a biholomorphic map. Set $a=f^{-1}(0)$. Then $h=f\circ g_a^{-1}$ is also a biholomorphic map $D\rightarrow D$ and we have

$$h(0)=f(g_a^{-1}(0))=f(f^{-1}(0))=0$$

By the lemma, $h(z)=\alpha z$ for some $\alpha\in\partial D$. But that means

$$f(z)=h\circ g_a(z)=\alpha \frac{z-a}{1-\overline{a}z}$$

which is the claim. $\square$

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  • $\begingroup$ you mean that the transformation is the form of $w=\alpha z$ where $|\alpha|=1$? $\endgroup$ – user114952 Jan 28 '14 at 22:37
  • $\begingroup$ @user114952 Yes that is also one. But you can also give a transformation that transports $a$ into $0$. It has to be of the form $\frac{z-a}{cz+d}$ right? So what can you plug in for $c,d$ to make $f$ stay inside the unit circle? $\endgroup$ – J.R. Jan 28 '14 at 22:45
  • $\begingroup$ @user114952 Ok, I will give you the complete solution, wait. $\endgroup$ – J.R. Jan 28 '14 at 22:55
  • $\begingroup$ So, $|z-a| \leq |cz+d|$ for $|z|=1$.. and how can I determine $c,d$? $\endgroup$ – user114952 Jan 28 '14 at 23:01

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