Möbius Transformation help

Question

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

Answer 1

Hint: the map is completely determined by its image on 3 points. So, compute the image of 3 points on the circle: $z=-1$, $z=i$ and $z=1$.

Answer 2

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

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#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 $x = 0$.

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#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.