Math 207 - Spring '17 - Fran¸coisMonard 1 Lecture 6 - Argument principle, Rouch´e'stheorem and consequences Material: [T, x4.4, 4.5], [G, xVIII.2]. [SS, Ch3, Sec. 4] Recall the Residue theorem, which allows to compute curve integrals of meromorphic functions in terms of their residues at the poles enclosed by that curve. Theorem 1 (Residue formula). Let f : U ! C meromorphic on U ⊂ C open with γ : I ! U a Jordan curve with Int (γ) ⊂ U, and such that f has no poles on γ(I). Then (i) Int (γ) encloses finitely many poles of f, call them z1; : : : ; zm. (ii) m Z X f(z) dz = 2πi Res (f; zj): γ j=1 We now show how this theorem can be used to compute the number of zeros and poles of f enclosed in a given region of the complex plane. When f is meromorphic, we have the following local representation of f: • near a zero or pole z0: there exists an integer k 6= 0, ρ > 0 and g analytic on Dρ(z0) (with k g nonvanishing near z0) such that f(z) = (z − z0) g(z) for z 2 Dρ(z0). Differentiating this equality, we obtain 0 k−1 k 0 f (z) = k(z − z0) g(z) + (z − z0) g (z); jz − z0j < ρ, so that 0 f k 0 (z) = g(z) + g (z); jz − z0j < ρ. f z − z0 f 0 • at any other point, f is analytic. f 0 As a conclusion, f is meromorphic wherever f is, and it has simple poles exactly located at the zeroes and poles of f, with residue the multiplicity of that zero/pole (with a negative sign if it is a pole). Combining this with the Residue theorem, we arrive at the next theorem. Theorem 2 (Argument principle). Let f : U ! C meromorphic on U ⊂ C open. Let γ : I ! U a Jordan curve with Int(γ) ⊂ U with no zeros or poles of f on γ(I). Then Z f 0(z) X f 0 dz = 2πi Res ; z = (2πi)(N − N ); f(z) f j 0 1 γ f0 zj poles of f where N0=N1 denote the numbers of zeros/poles of f inside γ, counted with multiplicity. (z−1)(z−i)3 R f 0(z) Example 1. Let f(z) = (z−2)5 . Compute γ f(z) dz for the following contours: (a) γ = @D1(−1); (b) γ = @D1:5(0); (c) γ = @D1:5(2): Math 207 - Spring '17 - Fran¸coisMonard 2 Theorem2 allows us to prove Rouch´e'stheorem. Theorem 3 (Rouch´e'stheorem). Let f; g : U ! C analytic on U ⊂ C open, γ : I ! U a Jordan curve with Int (γ) ⊂ U. Assume f has no zero on γ(I) and jf(z) − g(z)j ≤ jg(z)j; 8 z 2 γ(I): (1) Then f and g have the same number of zeros inside γ, counting multiplicities. You may think of f = (f − g) + g = "small" + "BIG" on γ(I) in the sense of (1): if f − g is \small" enough on γ(I), it does not perturb the number of zeros of g inside γ. f Proof of Rouch´e'stheorem. Due to (1), g has no zeros on γ(I) either. Define h := g , meromorphic on U, with no poles or zeros on γ(I). By Theorem2, we have Z h0(z) dz = 2πi(N0(h) − N1(h)) = 2πi(N0(f) − N0(g)); γ h(z) where the N's refer to numbers of zeros or poles inside γ. Using the equality above, the theorem will be proved if we can show that the left-hand-side is zero, which we now prove. R h0 To prove that γ h dz = 0, a way to rewrite (1) is, upon dividing by g, f(z) − 1 ≤ 1 8 z 2 γ(I) , jh(γ(t)) − 1j ≤ 1; 8 t 2 I; g(z) 1 so the closed curve h(γ(t)) is included in D1(1) − f0g, a convex set where z 7! z is analytic (that h cannot be zero on γ(I) comes from the fact that neither f nor g is vanishes on γ(I), as we stated above). Then Cauchy's theorem implies Z dz Z h0(γ(t)) Z h0(z) 0 = = γ0(t) dt = dz: h◦γ z I h(γ(t)) γ h(z) Hence the result. Example 2. Let f(z) = 1 + 2z + 7z2 + 3z5. Show that f has exactly two roots inside the unit disc. Answer: Apply Rouch´e'stheorem to g(z) = 7z2 and (f − g)(z) = 1 + 2z + 3z5. Among other consequences, Rouch´e'stheorem provides a short proof of the fundamental the- orem of algebra, with explicit bound on how large the roots are. In addition, we prove below a couple more consequences: the Open Mapping theorem and Hurwitz' theorem. Theorem 4 (Open Mapping Theorem). If f is holomorphic and non-constant in a region Ω, then it is open (i.e., f maps open sets to open sets). Proof. Suppose f non-constant. It is enough to show that if w0 2 f(z0) for some z0 2 Ω then there exists ρ > 0 such that Dρ(w0) ⊂ f(Ω). Indeed, consider the function g(z) = f(z) − w0. g is holomorphic, non-constant and has a zero at z0. By the local representation, there exists r > 0 such that g does not vanish on D_ (z ). In particular inf jg(z)j = C > 0. For w such that r 0 z2Cr=2(z0) jw − w0j < C, the function h(z) = f(z) − w satisfies jh(z) − g(z)j = jw − w0j < C ≤ jg(z)j; z 2 Cr=2(z0); so by Rouch´e'stheorem, h must have a zero inside Dr=2(z0), which means there exists z 2 Dr=2(z0) such that f(z) = w. That is, for every w 2 DC (w0), w belongs to f(Ω). Hence the proof. Math 207 - Spring '17 - Fran¸coisMonard 3 Hurwitz's theorem. Recall that a sequence fn of analytic functions fn : U ! C converges uniformly to f on K ⊂ U if and only if 8" > 0; 9n0; 8n ≥ n0; jfn(z) − f(z)j < "; 8 z 2 K: Recall that a uniform limit of analytic functions on K is itself analytic on K. For U ⊂ C an open domain, a sequence fn : U ! C of analytic functions converges normally to f on U if it converges uniformly to f on every compact subset of U. For the reason mentioned before this definition, the normal limit f as above is then analytic on U. Beware that normal convergence on U certainly does NOT imply uniform convergence on U. Examples: P n 1. The geometric series n≥0 z converges normally on the unit disk D1(0). (it's enough to show uniform convergence on every closed disk Dr(0) with r < 1). 2. The local power series of an analytic function converges normally to that function on its open disk of convergence. 3. The Laurent series of a function analytic on an annulus A(z0; r; R) converges normally to that function on the annulus. Theorem 5 (Hurwitz's theorem). Let U ⊂ C open and fn : U ! C a sequence of analytic functions on U, converging normally to a limit f on U. If z0 2 U is a zero of order m of f, there exists ρ > 0 and n0 2 N such that for every n ≥ n0, fn has exactly m zeros inside Dρ(z0), counting multiplicity. Moreover these zeros converge to z0 as n ! 1. m Proof. By local representation, there exists ρ > 0 such that f(z) = (z − z0) g(z) for every z 2 Dρ(z0), with g analytic on Dρ(z0) and g(z0) 6= 0. Upon shrinking ρ a bit, since g is continuous, we jg(z0)j can assume that jg(z)j ≥ 2 for every z 2 @Dρ(z0), so that, for jz − z0j = ρ, ρmjg(z )j jf(z)j = jz − z jmjg(z)j = ρmjg(z)j ≥ 0 ≡ c > 0: 0 2 Upon shrinking ρ again, we can assume that Dρ ⊂ U so that by normal convergence applied to the compact set K = Dρ(z0), there exists an integer n0 such that for n ≥ n0, maxz2K jfn(z)−f(z)j ≤ c. In particular, for every n ≥ n0 and jz − z0j = ρ, we have the estimate jfn(z) − f(z)j ≤ c ≤ jf(z)j: By Rouch´e'stheorem, this implies that fn has the same number of zeros as f inside Dρ(z0), counting multiplicity, i.e., m zeros. That these zeros converge to z0 is clear because we can shrink ρ as much we want, f will still have m zeros inside Dρ(z0), and following the proof above, for n large enough, fn much have its zeros inside Dρ(z0). Pn zk z Example: As a consequence, because the sequence fn(z) = k=0 k! converges normally to e on C, we can show that for every R > 0 there exists n0 such that for every n ≥ n0, fn has no zero inside DR(0). This is striking since fn has exactly n roots. This tells us that somehow, all the roots we add at every step should all escape at 1 ! Math 207 - Spring '17 - Fran¸coisMonard 4 References [G] Complex Analysis, Theodore W. Gamelin. Undergraduate Texts in Mathematics, Springer.1 [SS] Complex Analysis, Elias M. Stein and Rami Shakarchi, Princeton Lectures in Analysis II.1 [T] Complex Variables, Joseph L.