Lecture 6 - Argument Principle, Rouch´E’Stheorem and Consequences

Home , Argument principle, Theorem, Zeros and poles

Math 207 - Spring ’17 - Fran¸coisMonard 1

Lecture 6 - Argument principle, Rouch´e’stheorem and consequences

Material: [T, §4.4, 4.5], [G, §VIII.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 ∈ Dρ(z0). Diﬀerentiating this equality, we obtain

0 k−1 k 0 f (z) = k(z − z0) g(z) + (z − z0) g (z), |z − z0| < ρ,

so that 0 f k 0 (z) = g(z) + g (z), |z − z0| < 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 ∞ γ f0 zj poles of f where N0/N∞ 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 |f(z) − g(z)| ≤ |g(z)|, ∀ z ∈ 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) − N∞(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 ∀ z ∈ γ(I) ⇔ |h(γ(t)) − 1| ≤ 1, ∀ t ∈ I, g(z)

1 so the closed curve h(γ(t)) is included in D1(1) − {0}, 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 theorem 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 ∈ f(z0) for some z0 ∈ 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 |g(z)| = C > 0. For w such that r 0 z∈Cr/2(z0) |w − w0| < C, the function h(z) = f(z) − w satisﬁes

|h(z) − g(z)| = |w − w0| < C ≤ |g(z)|, z ∈ Cr/2(z0), so by Rouch´e’stheorem, h must have a zero inside Dr/2(z0), which means there exists z ∈ Dr/2(z0) such that f(z) = w. That is, for every w ∈ 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

∀ε > 0, ∃n0, ∀n ≥ n0, |fn(z) − f(z)| < ε, ∀ z ∈ 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 ∈ U is a zero of order m of f, there exists ρ > 0 and n0 ∈ 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 → ∞.

m Proof. By local representation, there exists ρ > 0 such that f(z) = (z − z0) g(z) for every z ∈ Dρ(z0), with g analytic on Dρ(z0) and g(z0) 6= 0. Upon shrinking ρ a bit, since g is continuous, we |g(z0)| can assume that |g(z)| ≥ 2 for every z ∈ ∂Dρ(z0), so that, for |z − z0| = ρ, ρm|g(z )| |f(z)| = |z − z |m|g(z)| = ρm|g(z)| ≥ 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, maxz∈K |fn(z)−f(z)| ≤ c. In particular, for every n ≥ n0 and |z − z0| = ρ, we have the estimate

|fn(z) − f(z)| ≤ c ≤ |f(z)|.

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 ∞ ! 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. Taylor. Pure and Applied Undergraduate Texts Vol. 16, 2011.