Find an analytic function that maps the disk $\{|z|<1\}$ onto the disk $\{|w-1|<1\}$ so that $w(0)=1/2$ and $w(1)=0$ Find an analytic function that maps the disk $\{|z|<1\}$ onto the disk $\{|w-1|<1\}$ so that $w(0)=1/2$ and $w(1)=0$ 


The 3 points theorem: Given 3 point $z_1, z_2, z_3 $ always map into 3 distinct point $w_1,w_2,w_3$ and only one linearly tranformation map $w=f(z)$ then 
$$\frac{w-w_1}{w-w_3} .\frac{w_2-w_3}{w_2-w_1}= \frac{z-z_1}{z-z_3} .\frac{z_2-z_3}{z_2-z_1}$$


I'm not sure I really know how to do this problem, all info I can get from the question is the domain is a unit circle with center at the origin and the image is also a unit circle with center at $(1.0)$. I also know that this function map 2 points $0 \to 1/2$ and $1 \to 0$.
How can I use the 3 points theorem with only 2 points? I can't just make up th ethird point, can I ?
 A: You can't "just make up" the third point, but Möbius transformations have a very helpful property that allow you to determine where a specific other point is mapped:
If $T$ is a Möbius transformation mapping the circle (or straight line) $C$ to the circle (or straight line) $K$, and $z,\, z^\ast$ are two points symmetric with respect to $C$, then $T(z)$ and $T(z^\ast)$ are symmetric with respect to $K$.
Now, for your desired map, you know $C$ and $K$, and you know one point $z$ (namely $0$) not lying on $C$ and where it is mapped (namely $T(z) = \frac{1}{2}$). So you need only find the point symmetric to $z$ with respect to $C$ and the point symmetric to $T(z)$ with respect to $K$, and then you have your three points.
A: We just need a holomorphic function $f:\mathbb{D}\to\mathbb{D}$ that is onto and maps $0$ to $-\frac{1}{2}$ and $1$ to $-1$, so that $1+f$ fulfills the given constraints. Now, from this related question, we know that:
$$ f(z) = e^{i\theta}\frac{z-a}{1-\bar{a} z} $$
for some $a\in\mathbb{D}$, hence:
$$ \frac{1}{2}=ae^{i\theta},\qquad -1=e^{i\theta}\frac{1-a}{1-\bar{a}},$$
so $\theta=\pi$ and $a=-\frac{1}{2}$ leads to:
$$ f(z) = - \frac{z+\frac{1}{2}}{1+\frac{z}{2}} = -\frac{2z+1}{2+z}$$
and the wanted analytic function is $\color{red}{-1+\frac{3}{2+z}}.$
