Resolution for Assignment 3 - Topics in Geometry | MATH 6490, Assignments of Geometry

Download Resolution for Assignment 3 - Topics in Geometry | MATH 6490 and more Assignments Geometry in PDF only on Docsity! Math 6490: Assignment 3 Answers First a few basic facts: M1: Passing through any three distinct points in the Riemann sphere, there is exactly one “circle” (which may be a straight line). M2: Möbius transformations are angle-preserving. M3: For any Möbius transformation corresponding to a real matrix m = [ a b c d ] , we have m(R̂) = R̂, where R̂ = R ∪∞. M4: For any Möbius transformation corresponding to a real matrix m = [ a b c d ] and any circle C which is symmetric about the real axis (i.e. C = C), we have m(C) is also symmetric about the real axis, since m(C) = m(C) = m(C). 1. First by induction we have mn = [ 1 0 n 1 ] for any integer n. Then mn(0) = 0 and mn(∞) = 1/n. By M3, m(R̂) = R̂. Since R̂ meets iR̂ at right angles, by M2 we must have C = m(iR̂) also meeting R̂ at right angles. That proves that [0, 1 n ] is a diameter of C. Hence, the center is 1 2n and the radius is 1 2|n| . Alternatively, by M4, C is symmetric about the real axis, and we reach the same conclusion. 2. For any z and for any integer n, we have T n(z)− 1 T n(z) + 1 = 2n z − 1 z + 1 . by simply iterating the defining relation for T . For example, since T 2(z) = T (T (z)), we have T 2(z)− 1 T 2(z) + 1 = 2 T (z)− 1 T (z) + 1 = 2 · 2 z − 1 z + 1 = 22 z − 1 z + 1 . From this, we conclude T n(−1) = −1, T n(1) = 1, the two fixed points. Also, at z = 0 we have T n(0)− 1 T n(0) + 1 = −2n. By solving this, we get T n(0) = 1−2 n 1+2n . A similar calculation shows that T n(∞) = 1+2n 1−2n (a) It is easy to see T corresponds to a real matrix. Then by M3 or M4, the image of the unit circle under T is symmetric about the real axis. Since it passes through ±1, it must be the unit circle again. 1