Given $f(z) = \dfrac{z-a}{1-\bar{a}z}$, with $|a|<1$.
I showed that if $|z|=1$, then $|f(z)|=1$; if $|z|<1$, then $|f(z)|<1$.
However, I am stuck at showing that the map $f$ is "onto".
Is there any elementary way of showing that this map is onto?
I looked at similar questions here, but they are only showing that $f$ is "into".
Thank you very much!
Note that $$f(z)=\frac{z-a}{1-\overline{a}z}$$ has inverse $$g(z)=\frac{z+a}{1+\overline{a}z}.$$
The algebra used in proving that $f$ is one-to-one (or "into") can also be used to show that $f$ is "onto". \begin{align} w & = \frac{z-a}{1 - \bar a z} \\[10pt] z & = \frac{w+a}{1+\bar a w} = \frac{w-b}{1-\bar b w} \end{align} The same argument that shows that if $z$ is on the unit circle then so is $w$ will show that if $w$ is on the unit circle, then so is $z$, because $b$, like $a$, is within the closed unit disk. Hence for every $w$, there is a suitable $z$, so $f$ is onto.
Hint: $$f(z) = \frac{z-a}{1-\bar{a}z} \iff z = \frac{f(z)+a}{1+\bar a f(z)}$$