cayley transform

**PRLM** · January 29th 2010, 07:00 PM

i am not sure it this is the right place for this question but could anyone show me the proof of cayley fransform where you show the map $f(z)=i \frac{1-z}{1+z}$ takes the set $D=\{z \in C : |z|<1 \}$ one to one onto the set $U=\{z \in C : IM(z) >0 \}$ ?

any help would be appreciated.

**Opalg** · January 31st 2010, 12:18 AM

Originally Posted by PRLM

i am not sure it this is the right place for this question but could anyone show me the proof of cayley fransform where you show the map $f(z)=i \frac{1-z}{1+z}$ takes the set $D=\{z \in C : |z|<1 \}$ one to one onto the set $U=\{z \in C : IM(z) >0 \}$ ?

If $w=i \frac{1-z}{1+z}$ then, by elementary algebra, $z = \frac{i-w}{i+w}$ , and conversely. Thus the map $z\mapsto w$ is invertible, and hence one-to-one, from $\mathbb{C}\setminus\{-1\}$ to $\mathbb{C}\setminus\{-i\}$ . Also, $|z|<1\ \Longleftrightarrow\ \Bigl|\frac{i-w}{i+w}\Bigr|<1\ \Longleftrightarrow\ |i-w|<|i+w|$ . That last condition says that the distance from w to i is less than the distance from w to –i, and that in turn is equivalent to $\text{Im}\,w>0$ .

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