Find a Mobius transformation sending a half circle with x intercepts -1 and 1 to a line x=1? I need to find a Mobius transformation that sends the unit circle centered at $(0,0)$ to the line $x=1$.
I know that a Mobius transformation is of the form $M(x)=\frac{ax+b}{cx+d}$ with $ad-bc>0$.
My intuition says that I should take $x=-1$ and send it to $M(x)=1$ and take $x=1$ and send it to $M(x)=∞$ (because we want a vertical line at $x=1$). Thus,
$1=\frac{-a+b}{-c+d}$ and $∞=\frac{a+b}{c+d}$
Then $c+d=0$.
Is my intuition correct and how should I proceed?
 A: A Möbius transformation is uniquely definied by three points and their images. Even with the choices you made this only amounts to two.
Once you have decided on that, you can follow e.g. this post of mine to compute the transformation, one dimension lower than in that post.
Or you simply set up a linear system of equations. That system of equations will be underdetermined because you can scale all of $a,b,c,d$ by a constant factor without changing the Möbius transformation they express. So pick a value for one of these (as long as it is nonzero) and all the others will follow.
If you go with the bare wording of the original task, i.e. the requirement of sending a given circle to a given line, then this does leave you with three real degrees of freedom. You could pick any three distinct points on the circle (e.g. $-1,1,i$) and map them to any three distinct points on the line in order to satisfy the requirement. You chose two, and that is fine if any solution will suffice. If you need to consider the space of all solutions, you should be more careful.
A: We require a function $M: \mathbb C \rightarrow \mathbb C$ of the form $M(z) = \frac{az+b}{cz+d}$ with the property that the unit circle be mapped by $M$ to the line $Re(z) = 1$. We choose three points on the unit circle $i,1,-i$ and choose mappings $ i \mapsto 1+i; \, 1 \mapsto1; \,-i \mapsto1-i$. Then we have the relations: 
$$1 = \frac{a+b}{c+d} \implies a+b = c+d$$
$$1+i = \frac{ai +b}{ci+d} \implies (c+d)i-c+d=ai+b$$
$$1-i = \frac{b-ai}{d-ci} \implies -(c+d)i -c+d = b-ai $$
Using your method of choice for solving linear systems, one arrives with the relations $b=0,\, c=\frac{a}{2},\,d=\frac{a}{2}$. Choosing $a =2$ we have the Moebius transformation  $M(z) = \frac{2z}{z+1}$
