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$f(z)= a\frac{z-z_0}{\overline{z_0}\cdot z-1}$, where $\left|a\right| = 1$ and $\left|z_0\right| < 1$ .

If $\left|z\right|= 1$, it is obvious that $\left|f(z)\right| = 1$, thus $f(z)$ maps unit circle to unit circle. If $z= z_0$, $f(z)= 0$.

So why does this transformation send $\left|z\right|<1$ to itself?

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Please see if the edit was what you intended for. – AlanH Mar 23 '13 at 18:53
Yes thanks !!!! – sarah Mar 23 '13 at 19:02
Maybe this help… – Cortizol Mar 23 '13 at 19:10
up vote 1 down vote accepted


$$|z|<1\;,\;|w|<1\;,\;\;|a|=1\Longrightarrow \left|\;a\frac{z-w}{\bar w z-1}\;\right|=\frac{|z-w|}{|\bar w z-1|}<1\iff$$ $$\iff |z-w|^2<|\bar wz-1|^2\;\;(**)$$

To make things now clearer, perhaps, put $\,z:=x+iy\;,\;\;w=a+bi\,$ , so

$$(**)\iff (x-a)^2+(y-b)^2<(ax+by-1)^2+(bx-ay)^2\iff $$

$$\iff x^2+y^2+a^2+b^2-\color{red}{2(ax+by)}<a^2x^2+b^2y^2+\color{green}{2abxy}-\color{red}{2(ax+by)}+1+b^2x^2+a^2y^2-\color{green}{2abxy}\iff$$

$$\iff (a^2+b^2-1)(x^2+y^2-1)>0$$

And since the last inequality is clearly true (why?) , then...

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I will denote by $U(1)=\{z | |z|=1 \}$.

Note first that

$$f(z)=f(y) \Leftrightarrow \frac{z-z_0}{\overline{z_0}\cdot z-1}=\frac{y-z_0}{\overline{z_0}\cdot y-1} \Leftrightarrow (z-z_0)(\overline{z_0}\cdot y-1)=(\overline{z_0}\cdot z-1)(y-z_0)$$

$$\Leftrightarrow y-z=z-y\Leftrightarrow y=z$$

This proves that $f$ is one to one (which is something you probably already know). You can also get from here that $f(U(1))=U(1)$ [before we only knew $f(U(1)) \subset U(1)$].

Now let $z$ be so that $|z| <1$. Then the segment $z_0z$ doesn't intersect $U(1)$ and hence $f(z_0z)$ cannot intersect $U(1)$.

But $f(z_0z)$ is a closed curve which starts at $0$ and ends at $f(z)$. Since this curve doesn't meet $U(1)$, it lies entirely inside $D=\{ z | |z| <1 \}$, thus $|f(z)| <1 \,.$

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