A conformal mapping of point A to point B in upper half plane Given arbitrary A,B in the upper half plane, how do you write a conformal bijection mapping point A to point B?
My first thought is to take the inverse of a mapping $f$ that would send A to the origin. Then consider a mapping $g$ of $0$ to B, and compose $g\circ f^{-1}$
My other thought is to consider construct circles in the UHP touching the origin, centered at A and B. I think (from a previous exercise) that a lune created by the overlap of the circles can be mapped (with $f(z)=\frac{1}{z}$) to a vertical bar, which can then be mapped by the exponential map to the entire upper half plane, but I'm not sure if this is the most effective (or even correct) method. 
 A: The Möbius transformations
$$z\longmapsto\frac{az+b}{cz+d}$$
with $a,b,c,d\in\mathbb{R}$ and $ad-bc=1$ are conformal bijective maps from the upper half plane to itself. They describe a holomorphic action of $SL(2)$ on $\mathbb{H}$, and it can be shown easily that this action is transitive. Namely, we have the map
$$z=x+iy\longmapsto\pmatrix{y^{\frac{1}{2}}&xy^{-\frac{1}{2}}\\0&y^{-\frac{1}{2}}}$$
giving us for any $z\in\mathbb{H}$ a Möbius transformation sending $i$ to $z$ itself.
Remark: In fact, as the isotropy subgroup of $i$ is easily proven to be $SO(2)$ (with a short calculation), we have an isomorphism
$$SL(2)/SO(2)\cong\mathbb{H}.$$
This is a special case of the isomorphism of Siegel upper half space $\mathcal{S}_n\cong Sp(n)/U(n)$.
Thus, given any two points $z_1,z_2\in\mathbb{H}$, simply take a Möbius transformation sending $z_1\mapsto i$ and compose it with a Möbius transformation sending $i\mapsto z_2$.
A: This answer is probably not as deep as Daniel's but I think it will do. You want to construct such conformal equivalence via a specific automorphism of the unit disk. It goes as follow:


*

*Construct a conformal map from $\mathbb{H} \rightarrow \mathbb{D}$ by defining $z\mapsto (z-A)/(z-\bar{A})$. Note that $A$ will be sent to $0$.

*Construct a conformal map from $\mathbb{D} \rightarrow \mathbb{H}$ by defining $z\mapsto (B-y\bar{B})/(1-y)$. Now $0$ will be sent to $B$.


The composition of these two maps will give you a desired conformal equivalence between the upper-half plane that sends $A$ to $B$.
A: Let $A= a_1 + a_2i, B =b_1 + b_2i.$ Then the map
$$f(z) =\frac{b_2}{a_2} z + b_1 - \frac{a_1b_2}{a_2}$$
is a conformal map of the upper half plane $U$ onto itself with $f(A) = B.$ (Note that $f$ is just a positive dilation followed by a horizontal translation, both maps being conformal self maps of $U.$)
