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Lecture 8 - the Extended Complex Plane Cˆ, Rational Functions, M¨Obius Transformations

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Lecture 8 - the Extended Complex Plane Cˆ, Rational Functions, M¨Obius Transformations Math 207 - Spring '17 - Fran¸coisMonard 1 Lecture 8 - The extended complex plane C^, rational functions, M¨obius transformations Material: [G]. [SS, Ch.3 Sec. 3] 1 The purpose of this lecture is to \compactify" C by adjoining to it a point at infinity , and to extend to concept of analyticity there. Let us first define: a neighborhood of infinity U is the complement of a closed, bounded set. A \basis of neighborhoods" is given by complements of closed disks of the form Uz0,ρ = C − Dρ(z0) = fjz − z0j > ρg; z0 2 C; ρ > 0: Definition 1. For U a nbhd of 1, the function f : U ! C has a limit at infinity iff there exists L 2 C such that for every " > 0, there exists R > 0 such that for any jzj > R, we have jf(z)−Lj < ". 1 We write limz!1 f(z) = L. Equivalently, limz!1 f(z) = L if and only if limz!0 f z = L. With this concept, the algebraic limit rules hold in the same way that they hold at finite points when limits are finite. 1 Example 1. • limz!1 z = 0. z2+1 1 • limz!1 (z−1)(3z+7) = 3 . z 1 • limz!1 e does not exist (this is because e z has an essential singularity at z = 0). A way 1 0 1 to prove this is that both sequences zn = 2nπi and zn = 2πi(n+1=2) converge to zero, while the 1 1 0 sequences e zn and e zn converge to different limits, 1 and 0 respectively. Definition 2. limz!z0 f(z) = 1 if for every M > 0, there exists ρ > 0 such that jz − z0j < ρ 1 implies jf(z)j > M. Equivalently, limz!z0 f(z) = 1 if and only if limz!z0 f(z) = 0. Combining both definition above, we can say that limz!1 f(z) = 1 iff for every M > 0, there exists R > 0 such that jzj > R implies jf(z)j > M. g(z) Example 2. • If f has a pole of order k > 0 at z0, f may be written as f(z) = k with (z−z0) k 1 (z−z0) 1 g(z0) 6= 0 and g analytic near z0. Then f(z) = g(z) so limz!z0 f(z) = 0, i.e., limz!z0 f(z) = 1. • For f any nonconstant polynomial, limz!1 f(z) = 1. Definition 3. The extended complex plane (a.k.a. Riemann sphere) is the topological space C^ = C [ f1g, where open sets are generated either by open disks Dρ(z0) or neighborhoods of infinity. 1 is now an abstract point which is no different than the other ones. We may study the 1 behavior of a function f : C ! C at 1 by studying the function f( z ) near z = 0. The question now is: when can such a function be extended as an analytic function f~ : C^ ! C^ ? 1 Adjoining one point to C is not the only way to compactify it, this is why this process is called one-point compactification. Math 207 - Spring '17 - Fran¸coisMonard 2 1 ^ Definition 4. f : C ! C is analytic at 1 iff the function w 7! f( w ) has a limit in C at w = 0 and 1 (i) if this limit is finite, f( w ) is analytic at w = 0, 1 (ii) if this limit is 1, 1 is analytic at w = 0. f( w ) 1−z Example 3. Show that f(z) = 1+z is analytic at 1 upon extending it with the value −1 there. 1 Definition 5. f : C ! C has a pole of order n at 1 if f( w ) has a pole of order n at w = 0 (we allow n = 0, which covers the case of a finite limit). ^ Another way of saying that f has a pole at 1 is to say that limz!1 f(z) exists in C (i.e. either as a finite number, or as 1). Theorem 1. f can be extended into an analytic function f~ : C^ ! C^ if and only if f is meromorphic on C and limz!1 f(z) exists, either as a complex number or 1. This extension is done as follows : when f is a meromorphic function on C such that limz!1 f(z) exists in C^, we define f~ : C^ ! C^: • At z 2 C where f is analytic, define f~(z) = f(z). • At z 2 C where f has a pole, define f~(z) = 1. • At 1, set f~(1) = limz!1 f(z). 0 Analytic functions on C^ define conformal maps near every point where f 6= 0. ^ ^ 1 Theorem 2. The map s : C ! C defined by s(z) = z with s(0) = 1 and s(1) = 0 is a conformal equivalence of C^ onto itself. Proof. Check that s is one to one and onto: this is because s ◦ s = Id. s is analytic at every 1 1 z 6= 0; 1. At z = 1, s( w ) = w is analytic at w = 0 so s is analytic at z = 1. At z = 0, s(z) = z is analytic at z = 0 so s is analytic at z = 0. Theorem 3. f : C^ ! C^ is analytic (or meromorphic) if and only if f is a rational function. Proof. We only prove ( =) ), the converse is left to the reader. The first claim is that f has finitely ^ many poles z1; : : : ; zn on C (denote (d1; : : : ; dn) their multiplicities). Indeed, since f has a limit in C 1 _ as z ! 1, then there exists r > 0 such that w 7! f( w ) has not poles on Dr(0) (you may split cases 1 according to whether 1 is a pole of f or not). In particular, f has no poles on fjzj > r ; z 2 Cg. Then since D 1 (0) is compact, f has finitely many poles there, so the first claim is proved. We r d dn then define g(z) := (z − z1) 1 ··· (z − zn) f(z), and we now claim that g is a polynomial. g has removable singularities at z1; : : : ; zn and is analytic elsewhere, so it's an entire function. Moreover 1 d since f has a pole at 1 (of possible order 0), there exists d ≥ 0 such that f( w )w is bounded near zero, or equivalently, f(z)z−d is bounded near 1. This implies that g(z)z−(d+d1+···+dn) is bounded near 1 and from a previous exercise, since g is entire, this implies that g is a polynomial of degree at most d + d + ··· + d . As a conclusion, f(z) = g(z) is a rational function. 1 n Qn dj j=1(z−zj ) Math 207 - Spring '17 - Fran¸coisMonard 3 2 2 2 3 The Riemann sphere Consider the sphere in S = fx1 + x2 + (x3 − 1=2) = 1=4g ⊂ R centered at (0; 0; 1=2), with North Pole N = (0; 0; 1). S is sitting on the plane fx3 = 0g ≡ C, and we can define the stereographic projection map, for M 2 S − fNg to be the unique intersection point between the line NM and the plane fx3 = 0g. This map is a homeomorphism: it is bijective, continuous with continuous inverse ρ−1. As the neighborhoods of 1 we defined earlier are mapped via ρ−1 to neighborhoods of the North Pole N, we may extend ρ : S − fNg ! C into a homeomorphismρ ~ : S ! C^ by declaringρ ~(N) = 1 and ρ~(M) = ρ(M) for any M 2 S − fNg. N M 0 ρ(M) fx3 = 0g Figure 1: Stereographic projection In that sense, we can assimilate C^ as a two-dimensional sphere. We will get other (visual) opportunities to see why in the future. An important property of C^ is that it is now a compact topological space, unlike C. In particular, every sequence in C^ has at least one accumulation point. This will imply that discrete subsets of C^ can only be finite, with consequences on analytic functions C^ ! C^ (e.g. the \identity" theorem), namely: Theorem 4 (Identity theorem on C^). If f; g : C^ ! C^ are analytic and agree on an infinite set, then they are identically equal. This implies in particular that such non-trivial functions can only have finitely many zeros and poles. Linear Fractional (a.k.a. M¨obius)transformations Definition 6. A Linear Fractional Transformation is a transformation of the form h(z) = az+b cz+d , with a; b; c; d 2 C such that ad − bc 6= 0. a b Each non-singular 2×2 complex matrix A = c d determines a linear fractional transformation az+b −1 φA(z) = cz+d . A straighforward calculation shows that φA ◦ φB = φAB and that φA = φA−1 , −1 1 d −b where AB is matrix multiplication and A = ad−bc −c a . In other words, the mapping A 7! φA is a \group homomorphism". Most importantly, this property makes it easier to compute the composition of two LFT's on the fly, by computing the product of the corresponding matrices. Math 207 - Spring '17 - Fran¸coisMonard 4 Example 4. • Affine transformations L(z) = az + b (c = 0, d = 1) requires a 6= 0. 1 • s(z) = z , i.e. (a; b; c; d) = (0; 1; 1; 0). Note that for every λ 6= 0, the transformations ha;b;c;d and hλa,λb,λc,λd define the same trans- formation. One could add the normalizing condition ad − bc = 1, though it is sometimes more readable to not normalize.
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