How to remember/rederive the isomorphisms from the half planes to the unit disc I know that $$z \mapsto \frac{z-i}{z+i}$$ maps the upper half plane to the unit disc, and $$z \mapsto \frac{z-1}{z+1}$$ maps the right half plane.  Is there an intuitive way to construct such maps from scratch?
 A: Yes. Any Mobius map is determined by three points (that is, if two maps agree at three points, then they are the same).  
Mobius maps send boundaries to boundaries, and interiors to interiors - that is, if you want to send the upper half plane to the unit disk, then you want to take its boundary (the real axis) and that to the boundary of the unit disc (the unit circle). You need to be careful, because you might accidentally send the upper half plane to the outside of the unit disc, but you can easily tweak the map to swap those if you need to. So, pick your favourite points on the real line (mine are $0,1$ and $\infty$), and map them to your favourite three points on the unit circle (mine are $1,-1$ and $i$).  
For example, in the maps you gave, you send $(0,1,\infty)$ to $(-1,-i,1)$, which thankfully lie on the unit circle. If we check some nice point in the upper half plane, say, $i$, we see that goes to $0$ which is in the disc - just as we wanted! It's slightly tricky to construct a mobius map that sends any triple to any triple, but it's certainly doable if that's what you want?  
EDIT: There is also a very cute way of finding a Mobius map that sends any three points to any three points. I will give a sketch here: the trick is to send everything via $(0,1,\infty)$ (this is a very nice triple). Suppose you want to send the triple $(x_{1},x_{2},x_{3})$ to $(y_{1},y_{2},y_{3})$. First, construct $f$ such that $f(x_{1},x_{2},x_{3})=(0,1,\infty)$, and then make $g$ with $g(0,1,\infty)=(y_{1},y_{2},y_{3})$. $f$ looks like this: $$f(z)=\frac{(z-x_{1})(x_{2}-x_{3})}{(z-x_{3})(x_{2}-x_{1})}$$
The $z-x_{1}$ ensures we are $0$ at $x_{1}$, the $z-x_{3}$ makes sure we are $\infty$ at $x_{3}$, and the other terms are a correction for $z=x_{2}$. Similarly, the inverse of $g$ is $$g^{-1}(z)=\frac{(z-y_{1})(y_{2}-y_{3})}{(z-y_{3})(y_{2}-y_{1})}$$
Now invert this (using the fact Mobius maps can be represented by matrices!) and the composition $gf$ is what you want.
A: I would just use the fact that a linear fractional transformation is determined by the images of three points. Suppose I want to map the disk of unit radius centered at $i$ to the half-plane $\operatorname{Re}\ge 1$.  The points $0$ and $i\pm 1$ are on the boundary of the disk and the points $1$ and $1\pm i$ are on the boundary of the half-plane.  If I map $0$ to $1$, and $i\pm1$ to $1\pm i$ but not necessarily respectively, then what happens?  Symmetry hints that $2i$ should get mapped to $\infty$, so we should have
$$
z\mapsto \frac{\cdots}{z-2i},
$$
but just working on the first three points, what do we get?  A little counterclockwise-oriented half circle centered at $0$ and in the upper half-plane, and hence in the disk, should map to a little counterclockwise-oriented half circle centered at $1$ and in the half-plane to the right of $1$.  That suggests (draw the picture)
\begin{align}
& i-1 \mapsto 1+i\quad\text{and}\quad 1+i\mapsto 1-i, \\
\text{and not } & i-1 \mapsto 1-i\quad\text{and}\quad 1+i\mapsto 1+i
\end{align}
because the latter alternative would map the inside of the disk to the half-plane $\operatorname{Re}\le1$.  So:
\begin{align}
0 & \mapsto 1 \\
i-1 & \mapsto 1+i \\
i+1 & \mapsto 1-i \\[4pt]
z & \mapsto \frac{az+b}{cz+d} = f(z) \tag1
\end{align}
Line $(1)$ tells us we have $0\mapsto \dfrac b d$, so we have $\dfrac b d = 1$  Hence
$$
f(z) = \frac{az+b}{cz+b}.
$$
Then since we want $f(i-1)=1+i$ we have
$$
1+i = f(i-1) = \frac{a(i-1)+b}{c(i-1)+b}
$$
$$
(1+i)(c(i-1)+b) = a(i-1)+b
$$
$$
-2c + b(1+i) = a(i-1)+b
$$
$$
-2c+ib = a(i-1)
$$
$$
c = \frac{a+i(b-a)} 2
$$
$$
f(z) = \frac{az+b}{\frac{a+i(b-a)} 2 z +b} = \frac{2az+2b}{\Big(a+i(b-a)\Big)z + 2b}.
$$
Since we want $f(i+1)=1-i$ we get
$$
1-i = f(i+1) = \frac{2a(i+1)+2b}{\Big(a+i(b-a)\Big)(i+1) + 2b} = \frac{2a(i+1)+2b}{2a+b(i-1)}
$$
$$
\Big( 2a+b(i-1) \Big)(1-i) = 2a(i+1)+2b
$$
$$
2a(1-i) + 2ib = 2a(i+1)+2b
$$
$$
a = b\frac{i+1}2
$$
Therefore
$$
f(z) = \frac{b(i+1)z + 2b}{2bz+2b} = \frac{(i+1)z+2}{2(z+1)} \tag{warning: see below}
$$
Clearly some detail has gone wrong here, but this method should do it.
A different approach is to use the initial suggestion that it's $\dfrac{\cdots}{z-2i}$.
If $f(z) = {az+b}{z-2i}$ and $f(0)=1$, then $f(z) = \dfrac{az-2i}{z-2i}$.  Then the problem is to find $a$.
We want $f(1+i) = 1-i$.  So
$$
1-i = f(1+i) = \frac{a(1+i)-2i}{(1+i)-2i}.
$$
$$
(1-i)^2 = a(1+i)-2i
$$
This leads to $a=0$ and we get $f(z) = \dfrac{-2i}{z-2i}$.
Then
$$
f(1+i) = \frac{-2i}{(1+i)-2i} = \frac{-2i}{1-i} = 1-i
$$
so this works.
A: For an isomorphism from upper half plane to the unit disk, suppose $i\mapsto0$. Then $-i\mapsto\infty$, since $i,-i$ are symmetric (w.r.t. the $x$-axis) and $0,\infty$ are symmetric (w.r.t. the unit circle). Thus $z\mapsto\frac{z-i}{z+i}$ is what you want.
