Author |
Topic: point of symmetry (Read 8129 times) |
|
trusure
Newbie
Gender: 
Posts: 24
|
 |
point of symmetry
« on: Apr 20th, 2009, 5:19pm » |
 Quote  Modify
|
while I was trying to find a Linear Fractional Transformation from the upper half plane to itself, with 0 mapped to 1 and i mapped to 2i, I used that the point of symmetric of i is -i mapped to -2i which is the point of symmetric for 2i,(using the symmetric principle), so now I have 3 pints mapped to 3 points, and using the cross ratio I found f(z)=2.5 *z +1 ! which is not correct ??
so, where is my mistake ??
can anyone help me ?
Thank you
|
| « Last Edit: Apr 20th, 2009, 5:20pm by trusure » |
 IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender: 
Posts: 1948
|
|
Re: point of symmetry
« Reply #1 on: Apr 20th, 2009, 8:11pm » |
Quote Modify
|
Probably not, if you don't show what you did. Solving
(z, 0; i, -i) = (f(z), 1; 2i, -2i)
for f(z) should work.
|
|
IP Logged |
|
|
|
trusure
Newbie
Gender:
Posts: 24
|
|
Re: point of symmetry
« Reply #2 on: Apr 20th, 2009, 8:56pm » |
Quote Modify
|
does it matter the order of the points while using the cross ratio ?!!
|
| « Last Edit: Apr 20th, 2009, 9:17pm by trusure » |
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: point of symmetry
« Reply #3 on: Apr 20th, 2009, 9:26pm » |
Quote Modify
|
No, as long as you put the points in the same order as their images. E.g.,
(-i, 0; z, i) = (-2i, 1; f(z); 2i)
would also work, but not
(z, 0; i, -i) = (f(z), 2i; -2i, 1)
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print