7.2 Conformal mappings

Transcription

1 7.2 Conformal mappings Let f be an analytic function. At points where f (z) 0 such a map has the remarkable property that it is conformal. This means that angle is preserved (in the sense that any 2 smooth curves intersecting at z are mapped into curves which intersect at the same angle at f(z), both in magnitude and sense). To see this we can argue loosely as follows: if δz is an increment in z then the correspondng increment δw in w = f(z) is given by so that δw f (z)δz arg δw arg f (z) + arg δz (arg f (z) is well-defined since f (z) 0 by assumption). Therefore δw is obtained by rotating δz through an angle arg f (z) (and dilating by a factor f (z) ). Consequently angle is preserved. This discussion can be tightened by starting from an arc z = z(t), in the z-plane. It maps to w = f(z(t)) in the w-plane and dw dt = f (z(t)) dz by the chain rule, etc.. dt That is, very locally at a point z 0, the effect of f is to 1. translate a tiny ball around z 0 to be centred at w 0 = f(z 0 ); 2. rotate the ball by Arg f (z 0 ); 3. scale the ball by f (z 0 ). a t b Example: Let s look at f(z) =z 2. Suppose first that z 0 = re iθ 0. Thenf (z 0 ) 0. Consider the circle of radius r and the ray at angle θ. These intersect at right angles at z 0. The images of these curves are the circle of radius r 2 and the ray at angle 2θ, which again intersect at right angles. Furthermore, it is easy to see that while f is not globally one-to-one, it is locally one-to-one. In particular, f maps the disk D(z 0, z 0 /2) bijectively onto its image (since if f(z 1 )=f(z 2 ) then z 1 = ±z 2, and if z 1 D(z 0, z 0 /2), then z 1 D(z 0, z 0 /2)). On the other hand, the rays at angles 0 and π/2 intersect at z 0 = 0, the point where f (z 0 ) = 0. The images of these rays under f are the rays at angles 0 and π, so in this case the angle is not preserved. Further, the every disk D(z 0,δ) is mapped in a two-to-one way onto a point in its image D(z 0,δ 2 ). Exercise: Try this with f(z) =z k. 58

2 7.3 The Inverse Function Theorem We have seen already that if f (z 0 ) 0 then f is locally one-to-one and hence locally invertible. The remaining issue to address is whether the inverse function is analytic. Theorem 41. (Inverse function theorem) Suppose f is analytic on a neighbourhood of z 0 = x 0 +iy 0 and that f (z 0 ) 0. Then f maps some neigbourhood of z 0 conformally and bijectively onto a neighbourhood of f(z 0 ). In particular, f 1 exists and is analytic on a neighbourhood of f(z 0 ). Proof. by Suppose that f = u + iv and as before consider the function F : R 2 R 2 defined F ( ) x = y ( ) u(x, y). v(x, y) Since f (z 0 ) 0, det(df) =u 2 x + u 2 y = v 2 x + v 2 y 0, and so the two dimensional real inverse function theorem says that F has a local differentiable inverse G = ( s t) defined on a neighbourhood of (u 0,v 0 )=F(x 0,y 0 ). Furthermore (omitting some of the function arguments) ( ) ( ) sx s Dg(u 0,v 0 )= y = Df 1 1 vy u = y. t x t y u 2 x + u 2 v y x u x The fact that u and v satisfy the Cauchy-Riemann equations immediately shows that s and t do too at this point. Note that since f (z) must be non-zero on a neighbourhood of z 0 and f is an open mapping, this says that that the components of G must be continuous and satisfy the Cauchy-Riemann equations on a neighbourhood of (u 0,v 0 ) and hence G is analytic. Of course, once we know that f 1 is analytic we necessarily have that (f 1 ) (w) = 1 f (f 1 (w)). The situation that you see when f(z) =z k is in fact canonical. The proof of the following is in Assignment 3. Theorem 42. Suppose f is analytic at a, and that f (a) =0and f 0. Then f maps a neighbourhood of a onto a neighbourhood of f(a), in an m-to-1 way, where m is the multiplicity of the zero of f(z) f(a). 59

3 Do we really care about maps being conformal? It turns out that conformal maps are extremely important in differential equations. Suppose that φ is a conformal bijection from Ω 1 to Ω 2. The bijection φ determines a linear transformation T φ (f) =f φ which maps, say, the vector space of analytic functions on Ω 2 to the vector space of analytic functions on Ω 1. These composition operators turn out to be very important objects in the interface between complex analysis and operator theory. The fact that φ is conformal is not so important if you think of T φ as mapping analytic functions to analytic functions. You could however, think of Ω 1 and Ω 2 as being subsets of R 2. Fact: A function u :Ω 2 R 2 R is harmonic if and only if T φ (u) =u φ is harmonic. In other words, if you are looking to solve the heat equation 2 u x u y 2 =0 on Ω 2, then you could solve this on a simpler region Ω 1 and then just translate the solution across. A big and important question therefore is the following: Question: ForwhichopensetsΩ 1 and Ω 2 in the plane is it possible to find a conformal bijection f mapping Ω 1 onto Ω 2? If such a map does exist, how much choice is there? A few facts are easy to check. There is no analytic bijection from C onto D. Such a function would be a bounded entire map and Liouville s Theorem rules that out. You obviously can t have one of the sets connected and the other not. Since one can work with the connected components, it will suffice to just look at connected sets Ω 1 and Ω 2. Example: You should get some feeling for some of the more standard transformations. For example 1. the map φ(z) = Log z maps an open sector Ω 1 = {z : t 1 < Arg z<t 2 } 60

4 conformally onto a strip Ω 2 = {z : t 1 < Im z<t 2 }. 2. the map φ(z) = z 1 z +1 maps the right half-plane conformally onto the unit disk. 3. To map a strip Ω 1 = {z :0< Im z<1} conformally onto the unit disk, you could first apply the linear map φ 1 (z) =πz π 2 with image Ω 2 = {z : π 2 < Im z<π 2 } then the map φ 2 (z) =e z with image the right half-plane, and the the map to end up on the disk. Example: What if φ(z) = z 1 z +1 Ω 1 = D, Ω 2 = {z C :1< z < 2}? Remember that the image of a simply connected set under a continuous map might fail to be simply connected. For example, the image of C under the map z e z is C \{0}. And the opposite can happen. There is however no continuous bijection between these sets. Suppose that φ :Ω 1 Ω 2 is such a map. Let Γ 2 denote the circle z =3/2 inω 2 and let Γ 1 = φ 1 (Γ 2 ) Ω 1. The curve Γ 1 can be continuously deformed to the single point 0 via a homotopic family of curves {Γ t 1} t [0,1]. This would imply that curves φ(γ t 1) were also homotopic to a point which isn t true. While we are discussing simple connectedness, in the complex plane, there is a slightly simpler way of describing such sets. 61

5 7.4 Analyticity on the Riemann sphere The complex plane is homeomorphic to the sphere in R 3 minus the north pole, via the stereographic projection map. Adding in this north pole corresponds to adding a point at in the complex plane to form C = C { }. The Riemann sphere C is a slightly different beast to C. For example C is a compact topological space rather than a noncompact one. Both C and C have metrics. Although these metrics are not equivalent, they do produce the same topology on C. The main difference is that one now how proper convergence to. A ball around the origin now looks like B M = { } {z C : z >M} and z n means that for all M there exists N such that for all n N, z n B M. Also C is no longer a field, since the operations z + =, z =, z 0, don t allow cancellation. The operations do however behave relatively well. Theorem 43. A subset Ω C is simply connected if and only if both Ω and C \ Ω are connected (in C). Proof. Omitted. Note that whether a set Ω is simply connected does not depend on whether you consider it as a subset of C or or C, so this gives a perhaps easier definition for subsets of the plane. Example: Ω=C \{0} is not simply connected as C \ Ω={0, } which is not connected. The Riemann sphere allows us to define away certain bad behaviour of functions. Functions can be naturally extended to C if lim z f(z) exists. That is, if there exists w C such that for every ball B around w there exists M such that if z >M then f(z) B. Example: (i) Let f(z) = 3z2 + z 2z 2 +8 Then f(z) 3 as z and so we would set f( ) = (ii) If f(z) =z 2 z, then we would set f( ) =. Check that you can fill in the details. 62

6 (iii) Let f(z) =e z. Then lim z f(z) doesn t exist! If lim z f(z) exists, then this procedure make f continuous at. Differentiation is slightly more complicated as one can surely not take The idea is to spin the sphere around! f f( + δ) f( ) ( ) = lim! δ 0 δ Suppose that Ω is an open subset of C which contains, that f :Ω C is analytic on Ω \ { } and continuous at. We shall say that f is analytic at if there is a ball centred at such that f(z) =f( )+ a 1 z + a , z >M. z2 Alternatively, consider the function F (z) =f( 1 ) (on the obvious domain). Then f is z analytic at if F is analytic at 0. Example: (i) Let f(z) = 3z2 + z 2z 2 +8.Then This has a Taylor series expansion F (z) = z 2 z 2 +8 = 3+z 2+8z. z 2 2 F (z) = z 6 z2 2 z 3 +24z 4 +8z 5 + O ( z 6) ) which could be transformed into a Laurent series expansion for f(z) around. (ii) The function f(z) =z can be extended to C by making f( ) =. Morally, you feel that f ( ) should be 1. However in this case F (z) = 1 z expansion. (iii) Let Then f( ) = a c f(z) = az + b cz + d. and f has a series expansion around : does not have a Taylor series f(z) = a c ad bc 1 c 2 z (ad bc)d 1 (ad bc)d2 1 + c 3 z2 c 4 z (Check this!) An automorphism is an invertible structure preserving bijection. In our context, an automorphism of C is an analytic bijection f : C C. Thusiff is an automorphism of C 63

7 then f is entire and f (z) is never zero. This is not enough however, since f(z) =e z is an entire function with nonvanishing derivative, but it is not a bijection. Let Aut(C) denote the set of all automorphisms of automorphisms of C. This is a group under composition, with identity e(z) =z. See if you can prove the following result. Theorem 44. Aut(C) ={z az + b : a 0, b C}. (This is sometimes called the parabolic group.) Question: What about automorphisms of the sphere? We say that a map f : C C is meromorphic if it is analytic except at a set of isolated poles in C. We say that f has a pole at if F (z) =f( 1 ) has a pole at zero. At these poles z f(z) =. Thus, for example, f(z) =z is meromorphic function with a pole at. Definition: An automorphism of the Riemann sphere C is a meromorphic bijection f : C C with a meromorphic inverse. The set Aut(C) of all automorphisms of the Riemann sphere forms a group under composition, with identity e(z) =z. It turns out that all such maps are of a special form. We ll need the following result. Lemma 45. Suppose that f : C C is meromorphic, and that f is entire on C. Then f is a polynomial. Proof. Let F (z) =f( 1 ). The F is either analytic at 0, in which case f is a bounded z entire function, and hence is constant, or else F has a pole, or order k say, at 0. That is lim z 0 z k f(z) F (z) exists and is non zero. This corresponds to lim being finite. But using z z k the Taylor expansion f(z) z k = 1 z k ( a0 + a 1 z + a 2 z ) = a 0 z k + a 1 z k a k + a k+1 z +... Write the right-hand side as g(z)+h(z) where h(z) is the entire part. As z the lefthand side converges and g(z) 0soh(z) converges too. But that can only happen if h is constant. Thus the Taylor series for f is finite and hence f is a polynomial. 64

8 8 Möbius or Fractional Linear Transformations Already in second year you had a decent look at Möbius transformations. These are maps of the form f(z) = az + b where ad bc and a, b, c, d C. cz + d The condition ad bc 0 ensures that the map is invertible. Indeed, using the notation from above f 1 (w) = dw b cw + a. (If ad = bc then f is constant!) Since f is one-to-one, at any point of C where it is finite, f (z) must be nonzero by Theorem 40. In fact f (z) = ad bc (cz + d) 2. Thus Möbius transformations are conformal, except perhaps at and d/c. Möbius transformations work naturally on the Riemann sphere C with the understanding that f( ) = a ( ) d c, f =. c (Of course, if c = 0 this needs to be interpreted as just that f( ) =.) Since these maps are meromorphic (on C) and invertible, with meromorphic inverse, we have shown the following: Theorem 46. Every Möbius transformation is an automorphism of C. All Möbius transformations are composites of (i) translations, z z + b, (ii) rotations, z e iθ z, (iii) homothetics (or dilations), z rz, r > 0, and (iv) inversions, z 1/z. This is obvious if either a or c = 0. Otherwise just use the fact that az + b cz + d = a c + 65 bc ad c cz + d.

9 Exercise: Prove that any composition of such maps (i) (iv) is a Möbius transformation. It is worthwhile thinking about what these maps do to the Riemann sphere. (ii) rotations, z e iθ z: this just spins C on its NS axis. (iii) homothetics, z rz, r > 0: moves lines of latitude further towards one of the poles (north if r>1 and south if r<1). (iv) inversions, z 1/z: reflects the sphere in the real axis. (Check this!) (i) translations, z z+b: this is the harder one to visualise. These moves points around the sphere, but keep the north pole fixed. There are some very nice resources for thinking about these things: The YouTube video Möbius Transformations Revealed. See also the article from the Notices of the American Mathematical Society by its creators, Douglas N. Arnold and Jonathan Rogness, available at: Clearly then the set G of all Möbius transformations forms a group under composition, with identity e(z) =z. In group theory, this is known as the projective linear group, or linear fractional group, PGL(2, C). It is a subgroup of Aut(C). It can also be realized as a quotient groups of the general linear group GL(2, C) of invertible 2 2 complex matrices. Lemma 47. The map Φ:GL(2, C) G, ( ) a b A = c d f A (z) = az + b cz + d is a group homomorphism with kernel {λi : λ C, λ 0}. Proof. First note that Φ is well-defined since the condition for A to be in GL(2, C) is exactly the condition for f A tobeamöbius transformation. Let ( ) ( ) a1 b A = 1 a2 b, B = 2 GL(2, C). c 1 d 1 c 2 d 2 66

10 Then it is just algebra to check that f AB (z) = (a 1a 2 + b 1 c 2 )z + a 1 b 2 + b 1 d 2 (c 1 a 2 + d 1 c 2 )z + c 1 b 2 + d 1 d 2 = f A (f B (z)) = (f A f B )(z). Finally A ker(φ) if f A (z) =e(z) =z. That is az + b cz + d = z. This can only happen if a = d 0 and c = b =0. One can check that any circle on the Riemann sphere corresponds to either a circle in the plane, or if the circle goes through the point, a straight line. In this context we shall just use the word circle for all of these. One of the classical theorems about Möbius transformations is that they map circles to circles (on the Riemann sphere). This can be proved using the following. Definition: Given 4 distinct points z 1,z 2,z 3,z 4 C the cross-ratio is defined to be X(z 1,z 2 ; z 3,z 4 )= z 1 z 3 z 1 z 4 z2 z 4 z 2 z 3. Lemma 48. The elements of G preserve cross-ratios. Proof. Every element of G is a composite of translations, rotations, dilations and inversions, and the it is an easy exercise to show that the cross-ratio is preserved by each of these maps. For example X( 1 z 1, 1 z 2 ; 1 z 3, 1 z 4 )= 1 z 1 1 z 3 1 z 1 1 z 4 = 1 z 1 z 3 1 z 1 z 4 1 z 2 1 z 4 1 z 2 1 z 3 1 z 2 z 4 1 z 2 z 3 = z 3 z 1 z4 z 2 z 4 z 1 z 3 z 2 = X(z 1,z 2 ; z 3,z 4 ). Lemma 49. Every element of G is determined by its values on any three distinct points, and, moreover, can be chosen to map any three distinct points to any other three given points. 67

11 Proof. than z 1,z 2,z 3 ) Suppose that f G maps z 1,z 2,z 3 to w 1,w 2,w 3 respectively. Then for all z (other X(z,z 1 ; z 2,z 3 )=X(f(z),w 1 ; w 2,w 3 ). That is f(z) w 2 w1 w 3 = z z 2 z1 z 3. f(z) w 3 w 1 w 2 z z 3 z 1 z 2 You can rearrange this to get a uniquely determined formula for f as a Möbius transformation. (This is what implies that a Möbius transformation maps circles to circles.) Theorem 50. Suppose that f is an automorphism of C. Then f is a Möbius transformation. Proof. Since f is a bijection on C, there is a unique point a = f 1 ( ). One can choose a Möbius transformation (such as g(z) =a + 1 ) which maps to a and hence z f = f g is an automorphism of C which maps to. Since no other point maps to, f is entire and hence by Lemma 45 it must be a polynomial. But it is also a bijection, and so it can only have degree one. Thus f = f g 1 is a composition of elements of G and hence is also in G. One additional ingredient that one might add to this is some topological structure. Suppose that f n (z) = a nz + b n, c n z + d n One might say that f n f if f(z) = az + b cz + d. a n a n d n = b n c n and similarly for the other three components. a ad bc Alternatively, you could give GL(2, C) a topology from a norm or metric on M 22 (C) and then look at the quotient topology. Questions of whether Aut(C) are connected or not are often important in applications. 68

12 Example: Can you find a continuous family of Möbius transformations f t such that f 0 (z) =z, f 1 (z) = 1 z? You could try except that when t = 1 2 f t (z) = (1 t)z + t tz +(1 t) this is not a Möbius transformation! Writing the maps as matrices, you can do this by linear interpolation between: f 0 = ( ) ( ) 1 i 0 1 ( ) 1 i 1 1 ( ) 1 i 1 0 ( ) 0 i 1 0 ( ) 0 1 = f It is worthwhile tracking what happens to the unit disk under this family of maps. At a certain point it needs to get turned inside out! This occurs as the image of the unit circle passes over the north pole. Exercise: Prove that G is connected. 69