MHB Mobius transformations - I don't get this example

[nacho-man]

Please refer to the attached image (sorry for the bad cropping, they were on separate pages)

I don't get what is meant by "two finite points''. Are these any two points which aren't equal to infinity?
Could i have chosen f(1) and f(2) ? what am i not understanding.
IF someone could explain this, that would be wonderful.

I also don't quite understand the reason of what they say afterwards, Why should the line have to be in either the left or right half plane? Is it because the interior of C must be one-to-one, and for that to occur, C must be a disk excluding the point zero?
Following from this, it's inverse will be a line which is either in the negative plane, or either in the positive?

Thanks!

 

[poissonspot]

I believe what they mean by finite points in this case is points on the circle that do not map to infinity. I think the idea is that since we know the circle is mapped to a line (from this being a Möbius transformation and the point $z=4$ being a pole of
$w=f(z)$), we can pick two points on the circle and consider their images and that will determine the line so long as we don't pick the point that maps to infinity. So $f(1)$ wouldn't be any good as it is not on the circle C.

To your second question, I think the gist of what the solution says is that the inside and outside of the circle must map to the left or right of the line. Checkout Example 3 on pg 13:

http://www.johno.dk/mathematics/moebius.pdf

 

[nacho-man]

I believe what they mean by finite points in this case is points on the circle that do not map to infinity. I think the idea is that since we know the circle is mapped to a line (from this being a Möbius transformation and the point $z=4$ being a pole of
$w=f(z)$), we can pick two points on the circle and consider their images and that will determine the line so long as we don't pick the point that maps to infinity. So $f(1)$ wouldn't be any good as it is not on the circle C.

To your second question, I think the gist of what the solution says is that the inside and outside of the circle must map to the left or right of the line. Checkout Example 3 on pg 13:

http://www.johno.dk/mathematics/moebius.pdf

I think I need to back track a little bit.

Circles can be mapped to either a line or circle.
In this case we know that it is a line, because the circle contains the pole at $z=4$, since we are working with a circle of radius 2, centered at $(2,0)$.

How is $f(1)$ not on the circle? Do you mean that we have to pick points ON the boundary of the circle, and that they can't be inside?

 

[poissonspot]

I think I need to back track a little bit.

Circles can be mapped to either a line or circle.
In this case we know that it is a line, because the circle contains the pole at $z=4$, since we are working with a circle of radius 2, centered at $(2,0)$.

How is $f(1)$ not on the circle? Do you mean that we have to pick points ON the boundary of the circle, and that they can't be inside?

That was exactly what I meant. remember we are first considering the mapping of the "boundary" of the circle, which as you said is a line. If we choose two points other than z=4 on the the boundary then and consider their mappings we should get two points on the line (the line in this case being the y-axis).