Conformal mappings from the unit disc to itself I am reviewing conformal mappings for a complex analysis exam, and I can't get my head around this question.
I have some $a\in D(0;1)$, and for $z\in D(0;1)$ let $$f(z)=\frac{a-z}{1-\overline{a} z}$$
I have successfully shown, using the fact that the derivative of a conformal map must be non-zero everywhere, that this map f is conformal. But how can I show, that this is a bijection from $D(0;1)$ to itself? Do I show that it maps the boundary of $D(0;1)$ onto the boundary of $D(0;1)$ and if so, how do I do this?
Any help really appreciated.
Thanks.
 A: If it's conformal, it's bijective, so it suffices to show that it sends $D\rightarrow D.$ If $z\in D$, then $$|f(z)|^2=\frac{|z|^2+|a|^2-z\bar{a}-\bar{z} a}{1+|z|^2|a|^2-z\bar{a}-\bar{z}a}.$$ Recall that $|z|^2<1.$ Multiplying this by $1-|a|^2,$ we get that $|z|^2-|a|^2|z|^2<1-|a|^2,$ which implies that $|f(z)|^2<1.$
EDIT: The phrasing in the question is somewhat confusing, with the statement saying that it's conformal, which seems to be assuming what we want to prove in some sense. I've shown that it maps $D$ to $D$. As @Maxim pointed out, I should say something about bijectivity. In this case, it is clear because $f$ is an involution from $D$ to $D$, and hence it's bijective. 
A: $f$ is a Möbius transformation.
Let's take three points on the boundary of $D(0,1)$.  Namely, $1,-1$ and $i$.
$f(1)=\dfrac {a-1}{1-\bar a}$.  By symmetry (since $1$ is on the $x$-axis), $\mid f(1)\mid=1$.
Similarly, $f(-1)=\dfrac {a+1}{1+\bar a}$, so $\mid f(-1)\mid=1$, (since $a$ and $\bar a$ are symmetric wrt $-1$).
Next $f(i)=\dfrac {a-i}{1-i\bar a}$.  Write $a=x+iy$.  Then $f(i)=\dfrac{x+(y-1)i}{1-y-xi}$.  Thus $$|f(i)|=\sqrt{\frac{x^2+(y-1)^2}{(1-y)^2+x^2}}=1$$
Thus $f$ maps $S^1$ onto $S^1$.
Finally note that $f(0)=a\in D$.  So $f$ maps $D$ onto $D$.
A: Lets show that for $a \in D(0,1)$ the function $f(z)=\frac{a-z}{1-\overline{a}z}$ is bijective straight from the definition.
$f$ is injective:  Suppose $f(z)=f(w)$ 
then $\frac{a-z}{1-\overline{a}z} =\frac{a-w}{1-\overline{a}w}$
So $(a-z)(1-\overline{a}w)= (a-w)(1-\overline{a}z)$
or $a - z - |a|^2w +\overline{a}zw=a-w-|a|^2z+\overline{a}zw$
which simplifies to $(z-w)+|a|^2(w-z)=0$
or equivalently, $(z-w)(1-|a|^2)=0 \implies z=w$ 
$f$ is surjective:  Suppose that $w \in D(0,1)$ then it is easy to show (see cmk's answer) that if $f(z) = w$ has unique solution in $D(0,1)$.
