Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.