Möbius Transformation help Hey guys I need help on these 2 questions that I am having trouble on.
1) Show that the Möbius transformation $z \rightarrow \frac{2}{1-z}$ sends the unit circle and the line $x = 1$ to the lines $x = 1$ and $x = 0$, respectively.
2) Now deduce from this that the non-Euclidean distance between the unit circle and the line $x = 1$ tends to zero as these non-Euclidean lines approach the x-axis.
I know that the non-euclidean distance between the lines $x=0$ and $x=1$ goes to zero as $y$ approachs $\infty$.
But I dont know how to deduce that the unit circle would go to $x=1$ and $x=1$ to $x=0$ and how would I deduce from this. Its from my old past tests and I am trying to practice but I got stuck on this question.
Please help out thank you

These are exercises 8.6.5, 8.6.6 from "The Four Pillars of Geometry." I'll use the terminology of the book.

#1:
Call the transform $\varphi(z) = \dfrac{2}{1-z}$.
We know that Möbius transformations send non-Euclidean lines (circles and lines) to non-Euclidean lines. If we calculate the image of three points from the unit circle, we can uniquely determine the image of the unit circle as a whole.
Pick three points from the unit circle, say $\{-1, 1, i\}$. We have:
$$
\varphi(-1) = 1,\quad \varphi(1) = \infty,\quad \varphi(i) = \frac{2}{1-i} = 1 + i
$$
All points lie on the line $x = 1$. Hence, the image of the unit circle is the line $x = 1$.
Similarly for the line $x = 1$, pick the points $\{1, \infty, i + 1\}$:
$$
\varphi(1) = \infty,\quad \varphi(\infty) = 0,\quad \varphi(1+i) = 2i
$$
All points lie on the line $x = 0$. Hence, the image of the line $x = 1$ is the line $x = 0$.

#2:
We will show that the non-Euclidean distance between the unit circle and line $x = 1$ approaches zero as both non-Euclidean lines approach the point $1$ on the real axis.
Consider a point $p_1$ on the unit circle approaching the point $1$, and another point $p_2$ on the line $x = 1$ approaching the point $1$. Their images under $\varphi$ both approach $\infty$ on the lines $x = 1$ and $x = 0$ respectively by #1. Since we know that the non-Euclidean distance is invariant under all Möbius transformations, and we also know that the distance between the lines $x = 0$ and $x = 1$ approaches $0$ as $y$ approaches $\infty$, it follows that the non-Euclidean distance between $p_1$ and $p_2$ approaches $0$ as they approach the $x$ axis.
