Solution for 18.112 Problem Set 3

Solution for 18.112 ps 3 1(Prob 2 on P83). Solution: First we will prove the following two lemmas. Lemma 1. Reflection carries circles to circles. Proof: Reflection is the composition of two maps: conjugation z 7→ z̄ and linear transformation z 7→ R2 /(z − ā) + a, both maps will carry circles to circles, thus reflection carries circles to circles. Lemma 2. Suppose circles C1 , C2 are both symmetric with respect to a line l, and C3 is the reflection image of C1 with respect to C2 . Then C3 is also symmetric with respect to the line l. (Compare this with the symmetry principle!) Proof: Without loss of generality, we can suppose l is the x−axis, and the center of C2 is the origin. Then z ∈ C1 and z ∗ ∈ C3 are related by z ∗ z¯ = R2 . So z̄ ∈ C1 =⇒ z¯∗ ∈ C3 , i.e. C3 is symmetric with respect to l. Now we return to our original problem. The symmetric point of z is z ∗ = 1/(z̄ − 2) + 2. • Reflect the imaginary axis. By lemma 1, the image is a circle. By lemma 2, the image circle is symmetric with respect to the x−axis. More over, 0∗ = 3/2, ∞∗ = 2, 1 both lie on the x−axis, so the image circle is � � � � �z − 7 � = 1 . � 4� 4 • Reflect the line x = y. Now the line l is x + y = 2, and both (1 + i)∗ = (3 + i)/2 and ∞∗ = 2 lie on l. So the image circle is � � √ � � �z − 7 + i � = 2 . � 4 � 4 • Reflect the circle |z| = 1. The line l is still the x−axis. By 1∗ = 1, (−1)∗ = 5/3, we get the image circle � � �z − � � 4 �� 1 = . 3� 3 2(Prob 3 on P88). Solution: We consider the number of fixed points of S. If S has more than 2 fixed points, then it has to be identity map, which is automatically elliptic. Moreover, by equation (13) on page 86, the fixed point will always exist (may be the point ∞). Now we suppose that S has two distinct fixed points a and b, then we have S(z) − a z−a =k . S(z) − b z−b Thus z−a S n (z) − a = n z−b S (z) − b S n−1 (z) − a = k n−1 S (z) − b S n−2 (z) − a = k 2 n−2 S (z) − b = ······ z−a = kn . z−b 2 So k n = 1, which implies |k| = 1 and S is elliptic. (If one of a, b, say a, is infinity, then S(z) − b = k(z − b). By the same way above, we see that |k| = 1 and S is elliptic.) At last suppose S has only one fixed point a. Let T z = 1/(z − a) + a, which maps a to ∞ and ∞ to a. Since T is one-to-one and onto, the map T ST −1 has only one fixed point T a = ∞. Thus T ST −1 z = cz + d for some c = 6 0. I claim that c = 1 in this case, otherwise f = d/(1 − c) is another fixed point of T ST −1 . Now z = T S n T −1 z = (T ST −1 )n z = z + nd, which implies d = 0. So T ST −1 = Id, and S = T −1 (Id)T = Id is elliptic. 3 3(Prob 5 on P88). Solution: First it is easy to check that two points a, b ∈ C corresponding to diamet­ rically opposite points on the Riemann sphere if and only if ab¯ = −1 (Check it!), thus b = −1/¯ a. Let T be a linear transformation which represent rotation T̃ of the Riemann sphere. Then T̃ has two fixed points, A, B, which are opposite points. Thus T has two fixed points, a and −1/ā. Now consider the C2 circles, |z − a| = c. |z + 1/z̄| Let Z be the point on the Riemann sphere corresponding to z, then � 2|z − a|/ (1 + |z|2 )(1 + |a|2 ) |Z − A| � = |Z − B| 2|z + 1/ā|/ (1 + |z|2 )(1 + |1/ā|2 ) � 1 + |1/ā|2 |z − a| � = |z + 1/z̄| 1 + |a|2 is constant, i.e. the C2 circles are mapped to the “latitudinal” circles on the Riemann sphere, which are invariant under the rotation T̃ . So the C2 circles are unchanged under the mapping T , which tells us that T is elliptic. On the other hand, suppose T is elliptic and has two fixed points a, −1/ā, then T maps each C2 circle into itself, and maps C1 circle to some C1′ , and the angle between C1 and C1′ is argk(See paragraph 3 on page 86). So T̃ maps “latitudinal” circle to itself, and maps “longitudinal” circles to another “longitudinal” circle by rotating angle argk since the stereographic projection is conformal. Thus T̃ acts on C2 circle (“latitudinal” circle) by rotating a fixed angle argk, which implies that T̃ is a rotation which fixes A, B. Thus all linear transformations which represent rotations of the Riemann sphere are exactly all elliptic linear transformations with fixed points a and −1/ā (when a = 0, let −1/ā = ∞). 4 4(Prob 2 on P108). Solution: � � 2π 1). xdz = (r cos t)(−r sin t + ir cos t)dt |z|=r 0 = � 2π (−r 2 cos t sin t + ir 2 cos2 t)dt 0 2 � 2π r = [− sin 2t + i(cos 2t + 1)]dt 2 0 r2 = × i × 2π 2 = πr 2 i. � � z + r 2 /z 2). xdz = dz 2 |z|=r |z|=r � � z r2 = dz + dz |z|=r 2 |z|=r 2z � r2 1 = dz 2 |z|=r z r2 × 2πi 2 = πr 2 i. = 5(Prob 4 on P108). Solution: � � |z − 1||dz| = |z|=1 2π |eit − 1||ieit |dt 0 � 2π 2π = � 2π = � � 2π = | cos t + i sin t − 1|dt 0 0 0 = 0 � √ (cos t − 1)2 + sin2 t 2 − 2 cos tdt � 4 sin2 (t/2)dt = −4 cos (t/2)|2π 0 = 8. 5