MATH Non-Euclidean Geometry Exercise Set #8 Solutions

Transcription

1 MATH 68-9 Non-Euliean Geometry Exerise Set #8 Let ( ab, :, ) Show that ( ab, :, ) an ( a b) to fin ( a, : b,, ) ( a, : b,, ) an ( a, : b, ) Sine ( ab, :, ) while Likewise,, we have ( a, : b, ) ( ab, :, ) b b ( )( b ) ( )( b ) ( ab, :, ) b b ( b)( ) ( )( b) ( )( b ) ( )( b ) b b ( )( b ) ( )( b ) ( )( b ) ( )( b ) ( )( b ) ab a b + ab + a + b ( )( b ) a b + a + b a a + b b ( )( b ) ( )( b) ( b)( ) ( )( b) ( a, : b, ), :, Use this information 4 by Davi C Royster Page Spring 4

2 Thus, we have ( ab, :, ) () ( ab, :, ) () ( a, : b, ) (3) So using Equations 3 an ( a, : b, ) (4) From Equations (4) an (3) ( a, : b, ) (5) Then from Equations (4) an (5) ( a, : b, ) (6) Suppose ( ab, :, ) ( a, : b, ) What is? From Equation (4) above we have: ( ab, :, ) ( a, : b, ) ( ) + ± i 3 3 Fin the isometry γ of H in GL (R) whih sens i to 4 + 3i an to 5 Does this isometry have any fixe points in H? We will ompose the two isometries that we always use: (, z a: b,) sens a to, b to an to Thus, we nee to know where a thir point is sent Sine i goes to 4 + 3i an goes to 5, then the Poinaré line through an i must go to the Poinaré line through 4+ 3i an 5 The line through an i is the irle entere at the origin of raius, so it goes through the ieal point Now, the istane of 4+ 3i from the origin is 5 so the line through 4+ 3i an 5 is the irle entere at the origin of raius 5 Hene, the isometry must sen i to 4+ 3i, to 5, an to 5 Now we are in the position to use our stanar isometry We will use τ represente by the z, i:, followe by the inverse of σ, whih is frational linear transformation ( ) represente by the frational linear transformation ( w,4 3 i: 5,5) τ σ i 4+ 3i by Davi C Royster Page Spring 4

3 This means that we nee to solve ( z, i:, ) ( w,4 3 i: 5,5) ( w,4+ 3 i: 5,5 ) ( z, i:, ) + for w ( w+ 5)(4+ 3i 5) ( z )(+ i) ( w 5)(4 + 3i+ 5) ( z+ )( + i) 5z w z Thus, we an represent γ by 5 γ z z To see if γ has any fixe points, we solve γ z z 5z z z + z 7z 7 i 7 + i z, 4 4 Sine the first solution has a negative imaginary part it is not in H, thus the fixe point is (7 + i ) 4 4 Fin an isometry γ of H in GL (R) whih sens i to 3i an to As in the previous problem, we nee to see where a thir point is sent uner γ We know that sine γ sens i to 3i an to, it must sen the line through i an to the line through 3i an The other enpoint of the line through i an is The Poinaré line through an 3i is the irle entere on the x-axis that passes through these points We nee to fin the enter of that irle so that we an fin the raius Fin the mipoint of the segment 3 between (,) an (,3), The slope of this line is 3 Thus, the perpeniular bisetor has slope /3 an passes through the mipoint Using this information we fin the equation of this line to be 3 y x+ 3 6y 9 x This line intersets the x-axis at x 4 Thus, the raius is 5 an so we now know that sine γ sens to, it must also sen to 9 Thus, γ is most easily foun by solving for w in the following ( ) z, i:, ( w,3 i:9, ) z ( w 9)(3i+ ) i ( w+ )(3i 9) 3z 8 w + 3z + 4 by Davi C Royster Page 3 Spring 4

4 3 8 Thus, γ is represente by the matrix γ z z 3 5 The map 5 γ 3 is a rotation of H What is the enter of this rotation? The enter of the rotation is the point fixe by the rotation So we nee to fin the point z so that z z γ z γ z z z 5 z z z+ 5 z i, + i At this point you might say, There are two fixe points, so it annot be a rotation But wait, one of these points is not in H The point i oes not lie in H Thus, the enter of the rotation is the point + i (,) 6 Fin an isometry γ GL (R) whih fixes i an sens to The isometry, γ, fixes i an hene is a rotation about i through some angle What is this angle? Again, we nee to see where a thir point will go We look at the line through i an Sine i is fixe an is sent to, γ must sen the line to the Poinaré line through i an But this is the irle of raius entere at the origin Thus, γ () So, what is the isometry? ( wi, :, ) ( zi, :,) ( w )( i+ ) z ( w+ )( i ) i z w z + γ z z Thus, from what we know about rotations in the Poinaré upper half-plane, we have that θ θ os sin γ z z θ θ sin os whih implies that os θ sin θ Thus, 4 by Davi C Royster Page 4 Spring 4

5 os θ θ sin θ tan θ 3π 5π, 4 4 Sine the sine is negative an the osine is positive, the angle must be 3π 7 Fin the refletion of + i through the line with enpoints an 5 Sine we have a refletion, we know that we an represent this refletion by f ( z) a γ ( z) ( z) Sine this is a refletion, we know that it sens the line to itself, so that f (5) 5 From f () we get ( ) a a+ b + a 4a+ b+ 4 f () From f (5) 5 we get a+ b+ 5 These two equations then give us that 6, or a 7 Substituting this into the first equation gives us that b So, 7 set an we get a7 an b Thus, the refletion is represente by γ 7 Note that e t 9 > Now, we nee to fin f ( + i) 7 f ( + i) + i 7 7( + i) i ( + i) ( ( )) 8 Fin a formula for the refletion through the line with enpoints an We will o this just as the above problem Represent the refletion by f ( z) a γ ( z) ( z) Thus, f () an f ( ) Start with f () an 4 by Davi C Royster Page 5 Spring 4

6 ( ) a a+ b + a b Sine f ( ) we get a+ b+ These two equations give us that a an b Thus, we an represent this by γ Thus, f( z) γ ( z) ( z) z 9 Fin a formula for the refletion through the line whih goes through 3i an + 4i Possibly the best way woul be to fin the enpoints ' of the line through these two points On the other han, sine the whole line is fixe, we know that the refletion will fix both of these points f (3 i) 3i: f ( + 4 i) + 4i: ( 3 i) 3i a 3ai + b 3i 3i + a 3ai + b 9 + 3ai b+ 9 ( (+ 4 i)) + 4i a a( + 4 i) + b + 4i ( + 4 i) + a a+ b Putting these two together gives us b 9 an a 4 Setting gives γ 4 Thus, 4 9 f ( z) γ ( z) ( z) 4 4z 9 z by Davi C Royster Page 6 Spring 4

7 Fin the Poinaré istane between P + 3i an Q 8+ 4i Write your answer in the form ln( ab ) where a an b are positive integers We nee to fin the points M an N where the Poinaré line intersets the x-axis We know the tehnique, so applying that we fin that the enter of the irle is x5 an it has raius 5 Thus, the enpoints are at an Thus, the Poinaré istane is ( P, Q) log( Q, P: M, N) log(8 + 4 i,+ 3 i:,) (8 + 4i )( + 3i ) log (8 4 i )( 3 i ) log 6 + i 3 Fin the Poinaré istane PQ (, ) between P i an Q Note that both P an Q lie on the unit irle, so that M an N Thus, PQ (, ) log( QPMN, :, ) + i 3 log, i :, log log by Davi C Royster Page 7 Spring 4