Möbius transformation of the right half plane

I want to find a Möbius transformation which sends the right plane to the disk $|z-1|<1$.

The given disk is the unit circle placed with center at (1,0i) on the complex plane. I take the values of $z_1=1$, $z_2=2$ and $z_3=3$ for the cross-ratio:

\begin{equation} \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \end{equation}

But when choosing these values, the onhy way I can come up to find respective w values s by plugging each z in the $|z-1|<1$. That is for $z_1=1$, would be $x=1$ in $\sqrt{(x-1)^2+(i^2y^2)}= 0$ Then for $z_2=2$, $w_2=1$, $z_3=3$, $w_3=2$. Plugging in the cross ratio, I get

\begin{equation} \frac{(w-0)(1-2)}{(w-2)(1)}=\frac{(z-1)(-1)}{(z-3)(1)} \end{equation}

which gives $w=z-1$. But this is totally wrong.

How do I find the correct values of $w$?

This may not be the same method as you are considering, but here goes:

Every Mobius transformation takes the form $$w=\frac{az+b}{cz+d}$$ for $a,b,c,d\in\mathbb{C}$, and where $z=x+iy$ and $w=u+iv$, $x,y,u,v\in\mathbb{R}$

To achieve the required transformation, choose the origin to remain invariant, and let $\infty$ be mapped to the other end of the diameter of the circle, i.e. to $w=2$.

Therefore we have $b=0$ and $\frac ac=2$.

Therefore we can simplify the transformation as $$w=\frac{2z}{z+e}$$ where $e=p+iq$

Rearranging this gives $$z=\frac{ew}{2-w}=\frac{(p+iq)(u+iv)}{2-(u+iv)}$$ $$=\frac{(pu+ipv+iqy-qv)}{(2-u)-iv}\cdot\frac{(2-u)+iv}{(2-u)+iv}$$

Therefore the real part of $z$ is $$\frac{(pu-qv)(2-u)-(pv+qu)v}{(2-u)^2+v^2}$$ For the right half-plane, this needs to be positive, so $$2pu-2qv-pu^2+quv-pv^2-quv>0$$ $$\implies pu^2+pv^2-2pu+2qu<0$$

Now choose $p=1$ and $q=0$, so the region is $$u^2+v^2-2u<0\implies |w-1|<1$$ as required.

So the transformation is $$w=\frac{2z}{z+1}$$