Equivalence class of similar triangles I am trying to understand this particular example in a book I am going through.
Consider the set of all equivalence classes of similar triangles , T. The book says that if I apply a similarity to expand and shrink a given triangle so that its vertices , say A and B ,  lie on points $ 0 $ and $1$ in the complex plane, and the vertex C is given by $z$ in the upper half plane $ H = \lbrace z=x+iy,y>0 \rbrace $, then an arbitrary point on H would represent an equivalence class of similar triangles, and so an element of T. But then the book goes on to say, if instead of A and B , I were to put B and C in the positions 0 and 1 respectively, then A would be at $ \frac{z}{1-z}$ while exchanging A and B would give C as $ 1- \bar{z} $.  
Thus T is obtained by identifying complex numbers in H which can be transformed into one another as above. The number of identified points can vary from 6 to just one for $ \frac{1+\sqrt(3i)}{2} $, which corresponds to the equilateral triangle. Ultimately this identification shows that T cannot be a manifold. 
Can anyone please explain the working of this example??
 A: The map $f(w)=(w-1)/(z-1)$ satisfies $f(1)=0$ and $f(z)=1$, and is a similarity. It also satisfies $f(0)=-1/(z-1)=1/(1-z)$, explaining the result of rotating the chosen points $A,B$ to the points $B,C$.
Also a diagram shows why $1-\bar{z}$ is located also in $H$ but reflected in the line $x=1/2$, justifying the reaqult of interchanging $A,B$.
Note these transforms can be combined, e.g. one can apply the above map $f$ once for the given $z$, and then (using the new $z$) applied again to get the case of $C,A$ as moved to $(0,1).$
A motivation for the transform $f(w)$: Note that $f(w)=(w-1)/(z-1)$ is a transform in which $w$ is the variable while $z$ is a fixed complex number. So maybe it should be written as $f_z(w)$ to emphasize dependency on $z$. We can then view the computation of $f_z(w)$ as: First subtract 1 from $w$ (this is a horizontal shift to the left by 1 of the plane), and then divide by $z-1$. Division by a fixed complex number $a$ can itself be thought of as the combination of: Divide by the length of $a$ (which is a stretch/shrink of the complex plane by the factor $1/|a|$), and then rotate through an angle which is the opposite of that of $a$ (by this I mean the angle formed by the vector $a$ and the positive axis, and since we're dividing the rotation becomes clockwise rather than counterclockwise).
Now see what happens in specifically computing $f_z(w)$ for the three points $0,1,z$ making up the triangle. The 1 gets moved left to $0$ by the left shift of one unit, and stays there for the next part since $0$ is fixed by rotations or stretch/shrinks. What about $f_z(z)$? First $z$ is moved left 1 unit to $z-1$. It is now at distance $|z-1|$ from the origin, and at whatever angle $z-1$ makes. We want the final position of $z$ to be at $1$. So it's clear we must divide $z-1$ by its length (to make it one unit long) and then rotate the result downward to lie along the $x$ axis. I think this explains how $z$ moves to $1$.
We now have the new locations for $(1,z)$ as $(0,1)$, and our transformation is a similarity since it's a combination of left shift, stretch/shrink, and rotation. So we want to see where the third point $0$ of the original triangle winds up, which we can compute by simply plugging in and getting $f_z(0)=(0-1)/(z-1)=1/(1-z).$
A sidenote: We can always agree that $0,1$ be the longest side, and $0,z$ the next longest. This way we can get one unique value of $z=x+iy$ for each triangle, namely that value satisfying $1/2 \le x <1$ and $0 < x^2+y^2 < 1$. This is the intersection of the region bounded by a chord of a circle and its circumference with the upper half plane, and the equilateral triangle ends up at the top vertex, with isosceles triangles corresponding to the boundaries of this "fundamental region" Any originally given point $z \in H$ may be transformed using $f(z)$ and the other map $g(z)=1-\bar{z}$ so as to give a unique point in this region.
So even though $T$ is not a manifold, it looks like after the identifications it is a manifold with boundary of some kind. 
