360
Set PMath 9
9
9.1
More reection and inversion
Poincar reection e
Let be the boundary circle of a Poincar model of the hyperbolic plane. e Let O be the center of . Let A and B be two P-points in this hyperbolic plane. Suppose A, B, and O are not collinear. 9.1.1 P-line (A, B)
Find the construction steps of the P-line on A and B following these steps: Find the circle on A and B that is orthogonal to . Call it . Find the arc on that is interior to . (*)
Hints: (i) The P-line lies on a circle that is on A and B and is orthogonal to . (ii) The segment connecting the centers of and meets in a point that is on the P-line. (iii) Show the end points of the arc as open circles. (iv) Before dening the macro, hide all the objects other than the three input objects (, A and B) and your three output objects, the arc and its end-points.
Get experience by experimenting with your macro by creating several lines. Construct some parallel lines. Construct several non-intersecting line pairs. 9.1.2 P-pbis (A, B)
Find the construction for a second P-line so that the P-reection with respect to it maps A to B and maps B to A. (*)
Hints: Since the P-reection maps A to B and B to A, we see that P-line(A,B) P-line(B,A) = P-line(A,B). Any circle that inverts A to B must have its center on (the Euclidean) line(A,B). There is another pair of points, U and V, on P-line(A,B) that, like A and B, must also change places. They give another lien on the center. Find a circle with such a center that is orthogonal to and hence also to . When you nd your construction, use Cabri tools to verify/conrm to your satisfaction that your work is right. answer Your will include the construction of the arc on that is inside .
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PMath 360
Set 9
9.2
9.2.1
Inversions in the complex plane
H1 , f1
Let be the circle in the complex plane with center 0 and radius 10. 9.2.1.1 Find H1 , the Hermitian matrix of . (*)
9.2.1.2 Use the numbers in the matrix H1 to nd a function f1 so that the the formula w = f1 (z) represents inversion with respect to . (*) 9.2.1.3 Verify that the xed points (**) of f1 are exactly the points that satisfy the equation determined by the matrix H1 . That is, z = f1 (z) (z, 1)H1 (, 1)t = 0. (*) z 9.2.2 H2 , f2
Let be circle with center (-4, -3) and radius 5. Let H2 be the matrix, (and f2 be the function) representing . Do the same three tasks for H2 and f2 that you did for H1 and f1 in 9.2.1. (*) 9.2.3 The product
9.2.3.1 Convince yourself that the product of the three inversions, rst with respect to , then with respect to , and then with respect to again, might be expected to be a reection with respect to a line. 9.2.4 9.2.4.1 (*) The composition f1 f2 f1 Find (and simplify) the composition f3 (z) = f1 (f2 (f1 (z))).
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PMath 360
Set 9
9.2.4.2 (*)
Use the function f3 to write the corresponding matrix H3 .
9.2.4.3 Fixed points points of a line. (*)
Verify that the xed points (**) of f3 are the 2008-07-04
(*) Items marked with an asterisk should be submitted for marking. (**) We say z is a xed point of f if and only if z = f (z). The reason is that we sometimes think of a function f as causing or representing motion. Something get moved from one place to another. If something does not move, it has a xed location.
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