Linear fractional transformation

[sweetvirgogirl]

sooo ...
i am kind of clueless about how to determine a linear fractional transformation for a circle that maps on to a line or vice versa ...


like i do *kinda* get how to map a circle on to a circle ... or a line on to a line ...

 

[Hurkyl]

Well, a line is just a circle that passes through the point at infinity! So...

 

[sweetvirgogirl]

Well, a line is just a circle that passes through the point at infinity! So...

the way prof showed it in class ... he said pick three points on the circle and send it to three points on the line (or vice versa if we are trying to map a line to a cirle) ... but i am not sure how do i know what those three points are going to map to ...

on a circle |z| = 1, we could send i to infinity and -i to infinity and then the third point? maybe i am taking a wrong approach ... or maybe the prof didnt mean to tell us to take this approach and i misunderstood him

 

[shmoe]

on a circle |z| = 1, we could send i to infinity and -i to infinity and then the third point? maybe i am taking a wrong approach ... or maybe the prof didnt mean to tell us to take this approach and i misunderstood him

A linear fractional transformation sends only one point to infinity, you can't send two there. You have to take 3 distinct points on your circle and map them to 3 distinct points on your line.

 

[sweetvirgogirl]

A linear fractional transformation sends only one point to infinity, you can't send two there. You have to take 3 distinct points on your circle and map them to 3 distinct points on your line.

oops typo ... i meant we could send i to infinity and -i to - infinity and then the third point?

(-infinity and infinity are different, right?)

 

[AKG]

No, infinity and -infinity are the same.

 

[sweetvirgogirl]

No, infinity and -infinity are the same.

woopsie ... let's restart then ...

I send i to infinity, -i to zero ... what about the third point?
it has to do something with fixing orientation, right? how does that work?

(like i know the third point can not be totally arbitrary like the first two points were, right?)

 

[shmoe]

The third point just has to be on the 'target' line. If you're going for the real axis, you could take the third point to be 1. Have you seen the cross ratio?

Different choices of the third point on the real axis will get different transformations, but will still send your circle to the real axis.

 

[sweetvirgogirl]

The third point just has to be on the 'target' line. If you're going for the real axis, you could take the third point to be 1. Have you seen the cross ratio?

Different choices of the third point on the real axis will get different transformations, but will still send your circle to the real axis.

no we haven't covered cross ratio in class yet.

so i *sorta* get how to send a circle to a line ... ...
the circle |z| = 1 and the line Re((1+i)w) = 0 ... so i sent 1 to infinity, -1 to 0 and i = -1 and i came up with i*(z+1)/(z-1) ...
the book has the answer as (1-i)*(z+1)/(z-1)
i want to make sure my answer is right ...

also... to send a line to a circle ... i pick three arbitrary distinct points on the line and send them to three distinct points on the circle?

also ... if you have free time and you don't mind talking to a stranger over yahoo/aim ... please let me know ... thanks! :)

 

[shmoe]

so i *sorta* get how to send a circle to a line ... ...
the circle |z| = 1 and the line Re((1+i)w) = 0 ... so i sent 1 to infinity, -1 to 0 and i = -1 and i came up with i*(z+1)/(z-1) ...
the book has the answer as (1-i)*(z+1)/(z-1)
i want to make sure my answer is right ...

-1 is not on that line.

also... to send a line to a circle ... i pick three arbitrary distinct points on the line and send them to three distinct points on the circle?

Yes.

also ... if you have free time and you don't mind talking to a stranger over yahoo/aim ... please let me know ... thanks! :)

I don't have any chat programs, sorry.

 

[sweetvirgogirl]

-1 is not on that line.

how exactly do you determine that? coz the way i saw it ... -1 was on the line ... maybe i need some rest

 

[AKG]

If -1 were on that line, then -1 would be one of the w such that Re((1+i)w) = 0, i.e. we would get:

Re((1+i)(-1)) = 0
Re(-1-i) = 0
-1 = 0

 

[qxred]

There is no linear fractional function that maps a line on to a circle, or a circle on to a line, (unless the circle is a point), since linear fractional function preserves convexity. i.e., image of any convex set is convex , and the inverse image of any convex set is convex. See Stephen Boyd, Lieven Vandenberghe: Convex Optimization, Page 39