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This is my start:

$f$ is the function from the open unit disc to R2

$f(z)$ is onto since for every $w$ in the codomain, there exists a $z$ such that $f(z)=w.$ Hence $w=\dfrac{z} {(1-|z|)}$, so by taking moduli:

$$|w|=|z|/(1-|z| )$$ $$|w|(1-|z| )=|z|$$ $$|w|-|z||w|=|z|$$ $$|w|=|z|+|z||w|$$ $$|w|=|z|(1+|w|)$$ $$|z|=|w|/(1+|w|)$$

Now where do I go? Thanks, FGH.

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You need to give the domain and the codomain; it's part of the definition of $f$. –  Xabier Domínguez May 1 '12 at 16:57
    
If you don't give the question some context, it will be hard for you to get an answer. –  Pedro Tamaroff May 1 '12 at 16:58
    
One last line to conclude the proof: Thus $1-|z|=1/(1+|w|)$ and $z=w(1-|z|)=w/(1+|w|)$. QED. –  Did May 1 '12 at 17:01
    
Sorry! The domain is the unit disk and the co-domain is R2. –  user30243 May 1 '12 at 17:11
    
Does my indication allow you to finish the proof? –  Did May 1 '12 at 18:23
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1 Answer 1

up vote 2 down vote accepted

I'm retaining as much as possible from your own wording; but note the differences!

The function $f(z):={z\over 1-|z|}$ is a function from the open unit disc $D$ to ${\mathbb C}$.

The function $f$ is onto if (not: "since") for every $w$ in the codomain ${\mathbb C}$, there exists a $z\in D$ such that $f(z)=w\ $, i.e., $${z\over 1-|z|}=w\ .\qquad(1)$$ So by taking moduli: $$|w|={|z|\over 1-|z| }$$ or $$|z|={|w|\over 1+|w|}\ .\qquad(2)$$ On the other hand, taking arguments in $(1)$ for a $z$ with $|z|<1$ we get $$\arg(z)=\arg(w)\ .\qquad(3)$$ Equations $(2)$ and $(3)$ together imply that a $z$ of the required kind would necessarily be given by $$z:={w\over 1+|w|}\ .$$ Now we have arrived at this result not by means of a general theory about such problems, but by means of an ad-hoc procedure. Therefore we have to check whether the $z$ we have found indeed fulfills the conditions $z\in D$ and $f(z)=w$. I leave this verification to you.

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