MAT389 Fall 2016, Problem Set 2

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1 MAT389 Fall 2016, Problem Set 2 Circles in the Riemann sphere Recall that the Riemann sphere is defined as the set Let P be the plane defined b Σ = { (a, b, c) R 3 a 2 + b 2 + c 2 = 1 } P = { (a, b, c) R 3 Aa + Bb + Cc = D } Recall that circles in a sphere can alwas be epressed as the intersection of the sphere with a plane. In the net two problems, we will see which planes in R 3 do intersect the Riemann sphere, and which ones do not. 2.1 Show that P passes through the North pole, N = (0, 0, 1), if and onl if C = D. P passes through the North pole if and onl if the point (0, 0, 1) satisfies the defining equation Aa + Bb + Cc = D, i.e., if and onl if C = D. 2.2 Prove that P Σ if and onl if A 2 + B 2 + C 2 D 2 as follows: (a) Convince ourself that P Σ if and onl if the point of P closest to the origin call it p satisfies d(0, p) 1, where d is the usual, Euclidean distance in R 3 and 0 R 3 means the origin. Hint: a picture should suffice. (b) Show that D d(0, p) = 2 A 2 + B 2 + C 2 Hint: recall from multivariable calculus that the vector A, B, C is normal to P, and that, in fact, p = (λa, λb, λc) for the appropriate value of λ. Notice that P Σ consists of a single point (p, in fact) if and onl if A 2 + B 2 + C 2 = D 2. Hence, if A 2 + B 2 + C 2 > D 2, then P Σ is an actual circle in Σ (that is, its radius is strictl positive). 1. If d(0, p) = 1, then P and Σ intersect onl at p. If d(0, p) > 1, P and Σ do not intersect, b the definition of p as the closest point to Σ. If d(0, p) < 1, it is visuall clear that P and Σ intersect, and that the do so in more than one point. If d(0, p) = 1, P is tangent to Σ at p.

2 2. Let p be the point on P such that 0p is orthogonal to P. We claim that p = p. If not, the line segment 0p is the hpotenuse of the right-angled triangle 0pp, so d(0, p ) < d(0, p), contradicting the definition of p. Hence, p = p. Since 0p is orthogonal to P, it is parallel to the normal vector to P. Hence, 0p = λ A, B, C for some λ R. Since p lies on P, λ must satisf A(λA) + B(λB) + C(λC) = D, so λ = D A 2 + B 2 + C 2 = d(0, p) 2 = 0p 2 = λ 2 (A 2 + B 2 + C 2 ) = D 2 A 2 + B 2 + C 2. The stereographic projection Remember that stereographic projection establishes a bijection between the Riemann sphere Σ and the etended comple plane Ĉ as follows: Σ (a, b, c) N = (0, 0, 1) π Ĉ a + ib 1 c Ĉ z = + i π 1 ( , Σ ) , N = (0, 0, 1) 2.3 Prove that the image of a line in Ĉ under π 1 is a circle on Σ. Hint: recall (Problem 1.19) that the equation of a line in the (etended) comple plane can be alwas be written as p + q + r = 0 with p and q not simultaneousl zero, or as βz + βz + γ = 0 for some β C and γ R. Use the eplicit epression of π to obtain an equation in terms of a, b and c. Let L be the line given b the equation p + q + r = 0 (properl speaking, L consists of the points in C satisfing the aforementioned equations, together with ). Substituting = a/(1 c) and = b/(1 c) we find a p 1 c + q b 1 c + r = 0. Clearing denominators and rearranging, we arrive at pa + qb rc = r. ( ) This is the equation of a plane in R 3, and the image of L under π 1 is the intersection of it with the Riemann sphere. In fact, Problem 2.2 implies that this image is indeed a circle in Σ, for ( ) is of the form Aa + Bb + Cc = D with since p 2 + q 2 > 0. A 2 + B 2 + C 2 = p 2 + q 2 + ( r) 2 > ( r) 2 = D 2,

3 2.4 Prove that the image of a circle in Ĉ under π 1 is a circle on Σ. Let C be the circle given b the equation (recall Problem 1.22) αz z + βz + βz + γ = 0 with α R, β C and γ R satisfing β 2 > αγ. Substituting z = (a + ib)/(1 c) we find α a2 + b 2 (1 c) 2 + β a + ib 1 c + β a ib 1 c + γ = 0. Using the equation a 2 + b 2 + c 2 = 1 of Σ, we can simplif the above to α 1 c2 2a Re β 2b Im β + + γ = 0 (1 c) 2 1 c and α 1 + c 2a Re β 2b Im β + + γ = 0 1 c 1 c Clearing denominators and rearranging, we arrive at (2 Re β)a (2 Im β)b + (α γ)c = (α + γ) ( ) Once again, this equation describes a plane in R 3, and the image of C under π 1 is the intersection of it with the Riemann sphere. We can appeal to Problem 2.2 to show that it is a circle: ( ) is of the form Aa + Bb + Cc = D with A 2 + B 2 + C 2 D 2 = 4(Re β) 2 + 4(Im β 2 ) + (α γ) 2 (α + γ) 2 = 4 ( β 2 αγ ) > Consider the Möbius transformation Möbius transformations z T (z) = az + b cz + d Verif that the inverse is again a Möbius transformation namel, the one given b z T 1 (z) = (that is, check that T T 1 = T 1 T = I). dz b cz + a We need to check that (T T 1 )(z) = z and (T 1 T )(z) = z. For the first identit, write ( ) dz b (T T 1 )(z) = T = a dz b cz+a + b (ad bc)z = = z cz + a + d ad bc For the second, c dz b cz+a ( ) az + b (T 1 T )(z) = T 1 = d az+b cz+d b cz + d c az+b cz+d (ad bc)z = = z + a ad bc

4 2.6 Show that an Möbius transformation of the form T (z) = e iθ z z 0 z z 0, θ R, Im z 0 > 0 sends the real line to the circle of radius 1. In other words, T (z) = 1 whenever Im z = 0. If Im z = 0, then z = z and z z 0 = z z 0 = z z 0. Since the modulus of a comple number is the same as that of its comple-conjugate, we have z z 0 = z z 0 = z z 0 = z z 0 z z 0 = 1 (see also Problem 1.15). Since e iθ = 1, it follows that T (z) = 1 whenever Im z = 0. The condition Im z 0 0 ensures that T is not constant (i.e., that the condition ad bc 0 in the definition of Möbius transformation is satisfied). Indeed, if Im z 0 = 0, then z z 0 = z z 0 (for an value of z!), and so T (z) = e iθ. v T (0) z 0 w = T (z) 0 = T (z 0) u Find a Möbius transformation D H that takes the origin to the point 3 + 2i. Hint: first, use the Möbius transformation z z + i iz + 1 to take D into H and the origin to i. Then find a Möbius transformation H H that takes i 3 + 2i. Recall that the Möbius transformation preserving the upper half-plane H are all of the form T (z) = az + b, a, b, c, d R, ad bc > 0 cz + d

5 Let T 1 be the Möbius transformation z Z = T 1 (z) = z+i iz+1, sending D onto H and 0 to i. Let T 2 be the Möbius transformation Z w = T 2 (Z) = 2Z + 3. Then T 2 preserves H, and T 2 (i) = 3 + 2i. Hence, the Möbius transformation sends D onto H and maps 0 to 3 + 2i. w = T (z) = (T 2 T 1 ) (z) = (2 + 3i)z + (3 + 2i) iz + 1 Y 0 Z = T 1 (z) i X w = (T 2 T 1 )(z) w = T 2 (Z) v 3 + 2i u Note: we did not justif that the transformation z z + i iz + 1 takes D into H. The easiest wa to prove that it does is to find its inverse and check that it takes H to D i.e., that it is of the form given in Problem 2.6. Another is to take three points on the unit circle (sa 1, i and 1) and see that their images all lie along the real ais. 2.8 Find a Möbius transformation D D that takes 1/2 to 1/3. Hint: recall that the Möbius transformations preserving the unit disc D take the form T (z) = e iθ z + α αz + 1

6 for some θ R and α C satisfing α < 1. Find two of these: one that takes 1/2 to 0, and another that takes 0 to 1/3. Note that T α : z z + α αz + 1 sends 0 α and α 0. Hence, T 1/2 sends 1/2 0, and T 1/3 sends 0 1/3. The composition T 1/3 T 1/2 preserves the unit disk, D, and sends 1/2 to 1/3. The eplicit computation ields ( T1/3 T 1/2 ) (z) = 1 5z z 5 Y 1/2 Z = T 1/2 (z) 0 X w = T 1/3 (Z) w = (T 1/3 T 1/2 )(z) v 1/3 u 2.9 Find the unique Möbius transformation taking i 1, 1 0, i 1.

7 Hint: recall that the unique Möbius transformation that takes z 1 0, z 2 1, z 3 is given b the cross-ratio T (z) = z z 1 z 2 z 3 z 2 z 1 z z 3 Note: this is the ver eample we did in class. See if ou can reproduce it without our class notes. Let z 1 = i, z 2 = 1, z 3 = i, and w 1 = 1, w 2 = 0, w 3 = 1, and define T and U as the unique Möbius transformations taking z 1 0, z 2 1, z 3, and w 1 0, w 2 1, w 3, respectivel. Eplicitl, Then T (z) = z + i 1 i 1 + i z i = 1 iz w , and U(w) = z i w 1 = w w. U(w) = T (z) w w = 1 iz z i w = i 1 z 1 + z. i Y 1 Z = T (z) 0 1 X i w = (U 1 T )(z) Z = U(w) v u Topolog in the comple plane

8 2.10 For each of the choices of S below, do the following: (a) classif all the points of C according to whether the are interior, eterior or boundar points of S; (b) decide whether S is open, closed, both open and closed, or neither open nor closed; (c) decide whether S is connected; (d) decide whether S is bounded. (i) S = {z C 0 < Arg z < π/2}; (ii) S = {z C z z 4 }; (iii) S = {z C 0 < z z 0 < δ}, where z 0 C and δ R >0 ; (iv) S = { z C Re(z 2 ) > 0 } {0}; (v) S = C. 1. (i) The interior of S is S itself, the open first quadrant. The boundar consists of the ras from 0 along the positive real ais and the positive imaginar ais, including the origin. The rest of the points of C are eterior to S. π/2 (ii) S is the set of points which are at least as far from 0 as the are from 4 (recall Problem 1.21) that is, S = {z C Re z 2}. Hence, the interior of S is {z C Re z > 2}, the eterior is {z C Re z < 2}, and the boundar is {z C Re z = 2}. 0 4

9 (iii) S is the open disk of radius δ and centre z 0, minus the point z 0. All points of S are interior. The boundar is the union of the one-point set {z 0 } with the circle {z C z z 0 = δ}. The eterior is the comple plane minus the closed disk {z C z z 0 δ}. δ z 0 (iv) Since Re(z 2 ) = 2 2, S consists of all the points between the lines = and = (not including the points on those lines), plus the point at the origin. All points of S ecept for the origin are interior points. The boundar consists of the two lines = and =. The rest of the comple plane makes up the eterior of S. (v) Ever point of C is interior to S. Consequentl, the boundar and eterior are empt. 2. (i), (iii) and (v) are open. (ii) and (v) are closed. (iv) is neither open nor closed. Note that (v) is both open and closed. 3. (i)-(v) are all connected. 4. Onl (iii) is bounded.

10 2.11 Find the accumulation points of each of the following subsets of C: (i) S = {i n n N}; (ii) S = {i n /n n N}; (iii) S = {z C 0 Arg z < π/2}; (iv) S = {( 1) n (1 + i)(n 1)/n n N}. (i) S has no accumulation points. i 1 1 i (ii) The onl accumulation point of S is the origin. n = 1 n = 2 n = 6 n = 3 n = 5 n = 4 (iii) The whole of the closed first quadrant {z C 0 Arg z π/2} {0} are accumulation points of S. π/2

11 (iv) S has eactl two accumulations points: 1 + i and 1 i. n = 2 n = 4 n = 1 n = 3 n = Prove that the interior of a subset S C, S = {z C z is an interior point of S}, is open, and that, in fact, it is the biggest open subset of C contained in S. Recall that a set is open if it does not contain an of its boundar points equivalentl, if all of its points are interior. If z S, there is an ɛ-neighborhood D ɛ (z) of z contained in S. We claim that, in fact, D ɛ (z) S which is the statement that z is an interior point of S. Indeed, let z 0 D ɛ (z), and denote δ = z 0 z. We then have that D ɛ δ (z 0 ) D ɛ (z) S, i.e., z 0 S. Now suppose that U S is open. If z U, there is an open neighborhoof D ɛ (z) U S, so z S. This shows that U S, and so S is the largest open set contained in S. D ɛ (z) ɛ z δ ɛ δ z 0 D ɛ δ (z)

12 2.13 Show that the intersection of finitel man open subsets of C is open. Hint: start b proving that the intersection of two open subsets of C is open, and then appl induction. Let U, V C be open sets, and z U V. Since z U and z V, there are neighborhoods D ɛ1 (z) U and D ɛ2 (z) V. Then D min(ɛ1,ɛ 2 )(z) U V, and it follows that U V is open. Now suppose that the intersection of k open sets is open, for some k 2. Let U 1,..., U k+1 C be open sets. B hpothesis, U 1 U 2 U k is open, so it follows that U 1 U 2 U k+1 = (U 1 U k ) U k+1 is open. Hence the intersection of an finite number of open sets of C is open. U D min(ɛ1,ɛ 2 )(z) V ɛ 2 z ɛ Consider the following famil of open subsets of C: { S n = z C z < }, n N n Is the intersection S = n N S n open? No: the intersection is the set {z C z 1}, which is the closed unit disk Show that the complement S c of an open subset S of C is closed, and viceversa. Use this and Problem 2.13 to conclude that the union of finitel man closed subsets of C is closed.

13 First, note that z S means that ever open disc centered at z contains a point in S and a point in S c. This is the same definition as for the boundar of S c, so it follows that S = S c. If S is open, then its boundar S is contained in the complement S c. B the observation in the last paragraph, S c contains all of its boundar points the ver definition of a closed set. On the other hand, if S is closed, it contains all of its boundar points, so S c contains none of them and is open. For the second part, let U 1,..., U n C be closed. Then U c 1,..., U c n are open and, b Problem 2.14, U c 1... U c n is also open. Therefore, U 1... U n = (U c 1... U c n) c is closed Prove that the closure of a subset S C, S = {z C z is an interior or boundar point of S} = S S = S S is closed, and that, in fact, it is the smallest closed subset of C containing S. Note: here S denotes the boundar of S. The first part of the statement follows from Problem 2.15 and the observation that the complement of S that, is the set of points eterior to S is an open set. Indeed, a point is eterior to S if and onl if it is interior to S c (just write out the definition!), and so (S) c = (S c ), which is open b Problem Since (S c ) is the largest open set contained in S c (once again, b Problem 2.12), taking complements sas that S is the smallest closed set containing S Show that S = S S c, and hence S is closed. B Problem 2.16, S = S S and S c = (S c ) S c. Recall from Problem 2.15 that S = S c. Then, S S c = ( S S) ( S c (S c ) ) = ( S S) ( S (S c ) ) = S ( S (S c ) ) = S = S In the last line, we have used that S and (S c ) are disjoint, since the first is the set of interior points of S and the second is the set of eterior points of S (see Problem 2.16).