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Riemann Mapping Theorem Chapter 7: Riemann Mapping Theorem Course 414, 2003–04 March 30, 2004 C 7.1 Theorem (Hurwitz’ Theorem). Let G be a connected open set in and (fn)n a sequence in H(G) which converges to f H(G) (uniformly on compact subsets of G). Suppose f 0, ¯ 2 6≡ D(a; R) G and f(z) is never zero on z a = R. Then there exist n0 such that for n n0, ⊂ j − j ≥ fn and f have the same number of zeros (counting multiplicities) in D(a; R). Proof. Since f(z) is never zero on the circle, we have inf f(z) = δ > 0: z a =R j j j − j For n large enough (say for n n0) ≥ sup fn(z) f(z) < δ=2 z a =R j − j j − j and thus on the circle z a = R we have j − j f(z) fn(z) < δ=2 < δ f(z) : j − j ≤ j j By Rouches´ theorem, fn and f have the same total number of zeros inside the circle z a = R j − j (counting multiplicity) for n n0. ≥ 7.2 Corollary. Let G be a connected open set in C and (fn) a sequence in H(G) such that each fn is never zero in G. Suppose fn f in H(G). If f(z) is ever zero in G, then f 0. ! ≡ Proof. This follows immediately from 7.1. If f is not identically zero but f(a) = 0, then by the identity theorem we can choose R > 0 sufficiently small that z = a is the only zero of f in the closed disk D¯(z; R) and also D¯(z; R) G. Counting multiplicity, f will have a positive number of zeros in D(a; R) and by Hurwitz so⊂ must fn for n large. 7.3 Definition. If G C is open and f : G C is an injective analytic function, then f is called a conformal mapping⊂ from G to f(G). ! 1 Recall that f(G) is necessarily open and f − : f(G) G is automatically analytic by the ! open mapping theorem. Also, f 0(z) is never zero in G and this leads to the angle-preserving property of conformal mapping that gives them their name: 1 2 414 2003–04 R. Timoney 1 If γ1 and γ2 are two C curves in G which meet at an angle θ to one another at a G (say γ1(0) = γ2(0) = a and arg(γ0 (0)) arg(γ0 (0)) = θ), then f γ1 and 2 1 − 2 ◦ f γ2 also meet at an angle θ at f(a). ◦ To avoid worrying about ambiguity of the argument, we could restate arg(γ0 (0)) arg(γ0 (0)) = θ 1 − 2 as iθ γ10 (0) e− > 0 γ20 (0) (with θ restricted to a range such as [0; 2π) — which means θ is the angle you would need to turn γ20 (0) anticlockwise to align it with γ10 (0)). The reasoning for the angle preserving property is that iθ (f γ1)0(0) iθ f 0(a)(γ1)0(0) iθ γ10 (0) e− ◦ = e− = e− > 0: (f γ2)0(0) f (a)(γ2)0(0) γ0 (0) ◦ 0 2 The term conformal really means angle-preserving at each point, but it is usual in complex analysis to use it for injective analytic functions. 7.4 Corollary. If G C is a connected open set and (fn) is a sequence of injective functions ⊂ n in H(G) such that fn f H(G), then the limit f is either constant or injective. ! 2 Proof. Suppose f is not constant and not injective. Then there exist a; b G, a = b, with 2 6 f(a) = f(b) = w. Taking fn(z) w and f(z) w instead of the original fn and f, we can assume that w = 0. − − We can find a positive δ < a b =2 so that D¯(a; δ) G, D¯(b; δ) G and f(z) never 0 on the circles z a = δ, z b j=−δ. Applyingj Hurwitz’⊂ theorem (7.1),⊂ we find that for all large j − j j − j n there exists an D(a; δ) with fn(an) = 0 = w. Similarly, for all large n there is bn D(b; δ) 2 2 with fn(bn) = 0 = w. By the choice of δ, D(a; δ) D(b; δ) = and therefore an = bn. Thus \ ; 6 fn(an) = fn(bn) = 0 and this contradicts injectivity of fn. 7.5 Examples. (i) The map f : z C: π=2 < (z) < π=2 z C: (z) > 0 f 2 − = g ! f 2 < g f(z) = ez is a conformal mapping (of its domain onto its target set, the right half plane). (ii) There is no conformal map φ of the unit disc D(0; 1) onto the whole complex plane C 1 (because the inverse function φ− : C D(0; 1) would be a bounded entire function which was not constant, contradicting Liouvilles! theorem). Chapter 7 — Riemann Mapping Theorem 3 (iii) If ad bc = 0, then the function − 6 az + b φ(z) = cz + d is called a Mobius¨ transformation or a linear fractional transformation. Since these maps are rational functions, we can (and often do) regard them as analytic functions φ: C^ C^: ! The condition ad bc = 0 is there to rule out constant functions and the ‘missing’ values are φ( ) = a=c,−φ( d=c6 ) = (unless c = 0, in which case a = 0 and d = 0, and we have φ1(z) = (a=d)z +− (b=d), φ1( ) = .) 6 6 1 1 7.6 Proposition. Let φ, be Mobius¨ transformations. Then (i) φ is a Mobius¨ transformation. ◦ ^ ^ 1 ^ ^ (ii) φ: C C is a bijection and its inverse φ− : C C is also a Mobius¨ transformation. ! ! (iii) The Mobius¨ transformations form a group (under composition). Proof. Exercise. 7.7 Proposition. Every Mobius¨ transformation can be expressed as a composition of Mobius¨ transformations of the following kinds: (i) z z + a (translation) 7! (ii) z λz (λ > 0) (dilation) 7! (iii) z eiαz (α R) (rotation) 7! 2 (iv) z 1 . 7! z Proof. If φ(z) = az+b and c = 0, we can write cz+d 6 a (cz + d) + b ad φ(z) = c − c cz + d a b ad 1 = + c c c2 d − z + c = φ5 φ4 φ3 φ2 φ1(z); ◦ ◦ ◦ ◦ 4 414 2003–04 R. Timoney where d φ (z) = z + 1 c 1 φ (z) = 2 z b ad φ3(z) = z c − c2 iα b ad φ4(z) = e z α = arg c − c2 a φ (z) = + z 5 c In the case c = 0, a iβ b φ(z) = (a=d)z + (b=d) = e z + ; d d with β = arg(a=d), and thus φ is a composition of a rotation, a dilation and a translation. 7.8 Proposition. Let P : C S2 denote the stereographic projection map from C to the Rie- mann sphere. If C is a circle! in C, then its image P (C) is a circle on S2. If L is a straight line in C, then P (L) (0; 0; 1) is also a circle on S2. [ f g 2 1 Conversely, if C1 is a circle on S , then P − (C1) is either a circle or a line in C. In other words, circles on the Riemann sphere correspond to circles and lines in the plane under stereographic projection (but the point at or the north pole has to be added to lines to close the circle on the sphere). 1 Proof. A circle (or line) in C has an equation A(x2 + y2) + 2βx + 2γy + C = 0 with A; β; γ; C R. In complex terms, this can be rewritten in the form 2 Azz¯ + Bz¯ + Bz¯ + C = 0: (B = β + iγ.) If A = 0 we can multiply across by A and rearrange this as 6 Az + B 2 = (Az + B)(Az¯ + B¯) = A2 z 2 + ABz¯ + ABz¯ + B 2 = B 2 AC j j j j j j j j − and the condition B 2 > AC corresponds to a positive radius (not an empty ‘circle’ or a single point). If A = 0, thenj j B 2 > AC = 0 is the condition for the equation to be that of a genuine line. j j Now ξ + iη P 1(ξ; η; ζ) = ; − 1 ζ − Chapter 7 — Riemann Mapping Theorem 5 and we can use this in the equation of the circle/line to say that the points (ξ; η; ζ) on the stereo- graphic projection of the circle/line onto the sphere must be those that satisfy (ξ + iη)(ξ iη) ξ + iη ξ iη A − + B¯ + B − + C = 0 (1 ζ)2 1 ζ 1 ζ − − − (and for lines the north pole P ( ) as well). Observe that 1 (ξ + iη)(ξ iη) ξ2 + η2 1 ζ2 1 + ζ − = = − = (1 ζ)2 (1 ζ)2 (1 ζ)2 1 ζ − − − − because ξ2 + η2 + ζ2 = 1. Using this, we can simplify the equation to get 2βξ + 2γη + (A C)ζ + (A + C) = 0: − This is the equation of a plane in R3, which intersects the sphere in a circle. Hence P maps circles and lines to circles. Conversely, starting with the equation of any plane in R3 a1ξ + a2η + a3ζ + c = 0: We can choose β = a1=2, γ = a2=2, A = (a3 + c)=2 and C = (c a3)=2. The point on the plane closest to the origin is − c (ξ; η; ζ) = 2 −2 2 (a1; a2; a3) a1 + a2 + a3 and this is inside the sphere exactly when a2 + a2 + a2 > c : q 1 2 3 j j So the condition for the plane to intersect the sphere in more than one point is c2 = (A + C)2 < a2 + a2 + a2 = 4β2 + 4γ2 + (A C)2 1 2 3 − or 4AC < 4(β2 + γ2): This is exactly what is needed to get a genuine circle or line.
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