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Prove that the function

$$ \begin{align} \phi (z) = i \dfrac{1 - z}{1 + z} \end{align} $$ maps the set $D = \{z \in \mathbb{C}: |z| < 1 \} $ one-to-one onto the set $U = \{ z \in \mathbb{C} : Im(z) > 0 \}$.

(This is exercise 1.9 in "Function Theory..." by Green & Krantz, and is also a claim on the Wikipedia page, though they have a function $W:U \to D $...)

For injective, I suppose that $$\begin{align} \phi (z) &= \phi (w) \\ \\ i \dfrac{1 - z}{1 + z}&= i \dfrac{1 - w}{1 + w} \end{align}$$ ... and after a few manipulations, end up with $z = w$.

For surjective, I am stuck.

Q: How to I prove "onto"? Do I work with $z, w$ in $(x + i \cdot y)$ form?

Also, how/where do I use that $|z| < 1$ (in $D$ ) or that $Im (\phi (z)) >0$ ?

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The Chaz 2.0
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    Notice this is a Möbius transformation, so with the matrix representation it's easy to obtain an inverse function. Prove that this inverse works when composed on either side of $\phi$. – Jose27 Jan 26 '12 at 18:05

2 Answers2

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Okay, since the comment is not really that helpful here's an approach: Take $w\in U$ and assume $w=\phi(z)$ for some $z\in \mathbb{C}$ then

$$i\frac{1-z}{1+z}=w$$

which implies

$$z=\frac{i-w}{i+w}$$

Now notice that $|z|<1$ if and only if $|i-w|^2<|i+w|^2$ now expand these in terms of the real and imaginary part of $w$ to obtain $(1-Im(w))^2<(1+Im(w))^2$ and this happens if and only if $Im(w)>0$. This implies that $\phi$ is onto.

Jose27
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Recall that $\mathrm{Im}(w)=(w-\bar{w})/(2i)$. So, with $r=|z|$, we have

$$\mathrm{Im}\;\phi(z)= \frac{1}{2}\left(\frac{1-z}{1+z}+\frac{1-\bar{z}}{1+\bar{z}}\right)=\frac{1-r^2}{|1+z|^2}>0,$$

which gives $r<1$. Additionally, with $w=u+iv$, solving the inequality

$$|\phi^{-1}(w)|^2=\left|\frac{i-w}{i+w}\right|^2=\frac{u^2+(1-v)^2}{u^2+(1+v)^2}<1$$

gives simply $v>0$. Bijectivity follows from the bidirectionality of $$\mathrm{Im}\;\phi(z)>0 \iff |\phi^{-1}(w)|<1.$$

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