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Solutions to Exercises Solutions to Exercises Solutions to Chapter 1 exercises: 1 i 1.1: Write x =Re(z)= 2 (z + z)andy =Im(z)=− 2 (z − z), so that 1 i 1 1 ax + by + c = a (z + z) − b (z − z)+c = (a − ib)z + (a + bi)z + c =0. 2 2 2 2 − a Note that the slope of this line is b , which is the quotient of the imaginary and real parts of the coefficient of z. 2 2 2 Given the circle (x − h) +(y − k) = r , set z0 = h + ik and rewrite the equation of the circle as 2 2 2 |z − z0| = zz − z0z − z0z + |z0| = r . 1.2: A and S1 are perpendicular if and only if their tangent lines at the point of intersection are perpendicular. Let x be a point of A ∩ S1, and consider the Euclidean triangle T with vertices 0, reiθ,andx. The sides of T joining x to the other two vertices are radii of A and S1,andsoA and S1 are perpendicular 1 if and only if the interior angle of T at x is 2 π, which occurs if and only if the Pythagorean theorem holds, which occurs if and only if s2 +12 = r2. 1.3: Let Lpq be the Euclidean line segment joining p and q. The midpoint of 1 Im(q)−Im(p) Lpq is 2 (p + q), and the slope of Lpq is m = Re(q)−Re(p) . The perpendicular 1 − 1 Re(p)−Re(q) bisector K of Lpq passes through 2 (p + q) and has slope m = Im(q)−Im(p) , and so K has the equation 1 Re(p) − Re(q) 1 y − (Im(p) + Im(q)) = x − (Re(p) + Re(q)) . 2 Im(q) − Im(p) 2 217 218 Hyperbolic Geometry The Euclidean centre c of A is the x-intercept of K,whichis 1 Im(q) − Im(p) c = − (Im(p) + Im(q)) 2 Re(p) − Re(q) 1 + (Re(p) + Re(q)) 2 1 (Im(p))2 − (Im(q))2 + (Re(p))2 − (Re(q))2 = 2 Re(p) − Re(q) 1 |p|2 −|q|2 = . 2 Re(p) − Re(q) The Euclidean radius of A is 2 2 1 |p| −|q| r = |c − p| = − p . 2 Re(p) − Re(q) 1.4: One hyperbolic line through i that is parallel to is the positive imaginary axis I = H∩{Re(z)=0}. To get a second hyperbolic line through i and parallel to , take any point x on R between 0 and 3, say x = 2, and consider the Euclidean circle A centred on R through 2 and i. 3 By Exercise 1.3, the Euclidean centre c of A is c = 4 and the Euclidean radius 5 of A is r = 4 . As the real part of every point on A is at most 2, the hyperbolic line H ∩ A is a hyperbolic line passing through i that is parallel to . 1.5: The Euclidean circle D√ through i and concentric to A has Euclidean centre −2 and Euclidean radius 5=|i − (−2)|, and so one hyperbolic line through i parallel to A is H ∩ D. To construct a second hyperbolic line through i parallel to , start by taking apointx on R between A and D,sayx = −4. Let E be the Euclidean circle centred on R passing through −4andi. By Exercise 1.3, the Euclidean centre 15 17 c of E is c = − 8 and the Euclidean radius is r = 8 . It is an easy calculation that the two Euclidean circles {z ∈ C ||z +2| =1} 15 17 and {z ∈ C | z + 8 = 8 } are disjoint, and so the hyperbolic line H ∩ E is a hyperbolic line passing through i that is parallel to . 1 1.6: If c = 0, then the Euclidean line Lc passing through i and 0 intersects S at ±i,andsoξ−1(0) = −i. Given a point c =0in R, the equation of the Euclidean line Lc passing through c and i is 1 1 y = − (x − c)=− x +1. c c Solutions to Exercises 219 To find where L intersects S1, we find the values of x for which c 1 |x + iy| = x + i − x +1 =1, c which simplifies to 1 2 x 1+ x − =0. c2 c As x = 0 corresponds to i,wehavethat 2c x = . c2 +1 So, 2c c2 − 1 ξ−1(c)= + i . c2 +1 c2 +1 1.7: Calculating, 2π 1 4π −1 ξ(1) = 1; ξ exp i = √ ;andξ exp i = √ . 3 3 − 2 3 3+2 1.8: Let z be a point of H. The Euclidean distance from z to R is Im(z). So, UIm(z)(z) is contained in H, but Uε(z) is not contained in H for any ε>Im(z). 1.9: Recall that K is bounded if and only if there exists some M>0sothat K is contained in UM (0). In particular, we have that UM (∞) is contained in X. For any z ∈ C − K, the Euclidean distance δ(z)fromz to K is positive, because K is closed, and so Uε(z) is contained in X for any 0 <ε<δ(z). Hence, the complement X of K in C is open. Suppose that W is an open subset of C.If∞ ∈ W ,thenW is contained in C, and by the definition of Uε(z), we have that W is open in C. If ∞∈W ,thenUε(∞) is contained in W for some ε>0. For this same choice of ε,wehavethatthecomplementY = C − W of W is contained in Uε(0), and so Y is bounded. The fact that Y is closed follows immediately from the fact that its complement C − Y = W is open in C. 1 ∈ 1.10: Given ε>0, we need to find N>0sothatzn = n Uε(0) for n>N. 1 1 Take N = ε . Then, for n>N,wehavethatzn = n <ε, as desired. Given ε>0, we need to find N>0sothatwn = n ∈ Uε(∞)forn>N. Take N = ε. Then, for n>N,wehavethatwn = n>ε,andsown ∈ Uε(∞), as desired. 1 ∈ ∩ 1.11: Note that 0 lies in X, because n Uε(0) X for every n>ε. However, there are no other points of X other than 0 and the points of X. 220 Hyperbolic Geometry If z ∈ C is any point with Im(z) =0,then UIm(z)(z) ∩ X = ∅. Also, because |x|≤1 for every x ∈ X, we see that U2(∞) ∩ X = ∅. ∈ R 1 ∈ Z−{ } If z is any point with Re(z) =0andRe(z) = n for n 0 , then either z lies between 1 and 1 for some m, p ∈ Z−{0},orelsez lies in one of the intervals m p ∞ −∞ − − 1 − 1 (1, )or( , 1). In the former case, let ε =min z m , z p ,sothat Uε(z) ∩ X = ∅. In the latter case, let ε = |z − 1| if z ∈ (1, ∞)orε = |z +1| if z ∈ (−∞, −1), so that Uε(z) ∩ X = ∅. Hence, X = X ∪{0}. For Y , take any z = x + yi∈ C. Given any ε>0, there exist rational numbers 1 1 a, b so that |x − a| < 2 ε and |y − b| < 2 ε, because Q is dense in R, a fact that can be proven by considering decimal expansions. Then, |(x + yi) − (a + bi)| <ε.Asforeachε>0 we can construct a point in Uε(z) ∩ Y ,wehavethateverypointofC lies in Y . As for any ε>0wehavethatn ∈ Uε(∞) for every integer n with n>ε,we also have that ∞∈Y . Hence, Y = C. 1.12: To show that X is closed in C, we show that C − X is open in C. Take z ∈ C − X. Suppose that for each ε>0, the intersection Uε(z) ∩ X = ∅.Foreachn ∈ N, choose some zn ∈ U1/n(z) ∩ X.Aszn ∈ X,thereissomexn ∈ X so that xn ∈ U1/n(zn) ∩ X. | − |≤| − | | − | 2 Combining these calculations yields that xn z xn zn + zn z < n . Hence, for each n ∈ N,wehavethatxn ∈ U2/n(z) ∩ X, which implies that z ∈ X. This result contradicts our original choice of z. 1.13: The Euclidean line LP can be expressed parametrically as N+t(P − N) = (tp1,tp2,tp3 +1− t) for t ∈ R. LP intersects the x1x2-plane when tp3 +1− t = 0, that is, when t = 1 . Hence, we see that 1−p3 p1 p2 ξ(P) = + i . 1 − p3 1 − p3 For ξ−1,letz = x + iy be any point of C, and note that z corresponds to the point Z =(x, y, 0) in R3.LetL be the Euclidean line between N and Z,and note that L is given parametrically by N+t(Z − N) = (tx, ty, 1 − t) Solutions to Exercises 221 for t ∈ R. To find where L intersects S2, we find the point on L whose distance from the origin is 1, which involves solving (tx)2 +(ty)2 +(1− t)2 = t2|z|2 + t2 − 2t +1=1 2 for t. There are two solutions: t = 0, which corresponds to N, and t = 1+|z|2 . The latter value of t yields 2Re(z) 2Im(z) |z|2 − 1 ξ−1(z)= , , . |z|2 +1 |z|2 +1 |z|2 +1 1.14: Write n g(z)=anz + ···+ a1z + a0, for n ≥ 1withan = 0. We need to quantify the statement that, if |z| is large, then |g(z)| is large. If we wanted to be precise, we could proceed as follows. By the triangle inequality, n n−1 |g(z)|≥||anz |−|an−1z + ···+ a1z + a0||. So, set A =max{|an−1|,...,|a0|} and note that n−1 n−1 n−1 |an−1z + ···+ a1z + a0|≤A (|z | + ···+ |z| +1)≤ nA|z| for |z|≥1.
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