The principle of the argument

If f is a non-zero meromorphic function on a domain U , then so are f^ and f^/f . Where are the poles of f^/f , and what are their residues?

Proposition 23.6.1 Suppose that f is a non-constant meromorphic func-tion on a domain U with singular set Sf and zero set Zf. The function f^/f is a meromorphic function on U with singular set Sf ∪ Zf. If f has a zero of order k at z₀ ∈ Zf then f^/f has a simple pole at z₀ with residue k. If f has a pole of order k at z₀ ∈ Sf then f^/f has a simple pole at z₀ with residue −k.

Proof The function f^/f is defined on U \ (Sf ∪ Zf) and is holomorphic there. If f has a zero of order k at z0 ∈ Zf, then f (z) = (z− z0)^kg(z), where g is a holomorphic function on U\ Sf and g(z₀)= 0. Then

f^(z) = k(z− z0)^k−1g(z) + (z− z0)^kg^(z).

Thus f^(z)

f (z) = k z− z0

+g^(z)

g(z) for z∈ U \ (Sf ∪ Zf).

Since g^/g is holomorphic in a neighbourhood of z₀, it follows that f^/f has a simple pole at z₀ with residue k.

The argument for a pole is very similar. If f has a pole of order k at z₀ ∈ Zf, then f (z) = (z− z0)^−kh(z), where h is a holomorphic function on U \ Sf for which h(z0)= 0. Then

f^(z) =−k(z − z0)^−k−1h(z) + (z− z0)^−kh^(z).

Thus f^(z)

f (z) = −k z− z0

+h^(z)

h(z) for z∈ U \ (Sf ∪ Zf).

Since h^/h is holomorphic in a neighbourhood of z0, it follows that f^/f has a simple pole at z0 with residue−k.

If z0 ∈ U \ (Sf ∪ Zf) then f (z0) = 0, and f^/f is holomorphic in a

neighbourhood of z₀. 2

We use this to prove the principle of the argument.

23.6 The principle of the argument 725 Theorem 23.6.2 (The principle of the argument) Suppose that f is a meromorphic function on a domain U with zero set Z_f and singular set S_f, and suppose that Γ is a cycle in U\ (Zf ∪ Sf) such that n(Γ, w) = 0 for Let us apply Proposition21.1.8.

Corollary 23.6.3 (Rouch´e’s theorem) Suppose that g is a meromorphic function on U for which S_g∩ [Γ] = ∅ and

Proof The conditions imply that f and g have no zeros and no poles in [Γ]. Applying Proposition 21.1.8 to each of the paths in Γ, we see that

n(f◦ Γ, 0) = n(g ◦ Γ, 0). 2

Rouch´e’s theorem is usually stated (and used) with the stronger condition that 0 <|f(w) − g(w)| < |f(w)|, for w ∈ [Γ].

These results can be used to locate the zeros of holomorphic functions.

We shall give some examples in Section 23.7.

We use the principle of the argument to consider the behaviour of a holomorphic function at a point where the derivative may be 0.

Theorem 23.6.4 Suppose that f is a non-constant holomorphic function on a domain U and that z₀ ∈ U. Let d be the least positive integer such that f^(d)(z0) = 0. Then there exist ρ > 0 and r > 0 such that for each w ∈ Nρ^∗(f (z0)) there exist exactly d points in N_r^∗(z0) satisfying f (z) = w.

These points are simple zeros of the function f− w.

Proof Since the zeros of f− f(z0) and f^ are isolated, there exists r > 0 such that M_r(z₀) ⊆ U and such that f(z) − f(z0) = 0 and f^(z) = 0 for z ∈ Mr^∗(z₀). Then f (z) = f(z0) for z ∈ [κr(z₀)], and n(κ_r(z₀), z) = 1 for z ∈ Nr(z₀). By the principle of the argument, n(f ◦ κr(z₀), f (z₀)) = d. Let V be the connected component of C\ f([κr(z₀)]) to which f (z₀) belongs.

Since V is open, there exists ρ > 0 such that N_ρ(f (z₀)) ⊆ V . Since the winding number n(f ◦ κr(z₀), w) is constant on V , n(f ◦ κr(z₀), w) = d, for w ∈ Nρ(f (z₀)), and so f − w has d zeros, counted according to mul-tiplicity, in N_r^∗(z0). Since f^(z) = 0 for z ∈ Nr^∗(z0), each of these zeros is a simple zero, and so there are d distinct solutions to the equation

f (z) = w in N_r^∗(z0). 2

We use this to describe the behaviour of a meromorphic function near a pole.

Corollary 23.6.5 Suppose that f is a meromorphic function on a domain U , with a pole of order k at z₀. Then there exist R > 0 and r > 0 such that if |ζ| > R there exist exactly d points in Nr^∗(z) satisfying f (z) = ζ. These points are simple zeros of the function f − ζ.

Proof There exists δ > 0 such that f (z) = g(z)/(z− z0)^k for z∈ N_δ^∗(z₀), where g is a holomorphic function on N_δ(z0) with no zeros in N_δ(z0). Let h(z) = (z−z0)^k/g(z) for z∈ Nδ(z0). Then h has a zero of order k at z0, and so there exist ρ > 0 and 0 < r≤ δ such that the conclusions of the theorem hold (with h in place of f ). Let R = 1/ρ. If|ζ| > R, then 1/ζ ∈ Nρ^∗(0), and there exist exactly d points in N_r^∗(z0) satisfying h(z) = 1/ζ, and these points are simple zeros of the function h−1/ζ. Since h(z) = 1/f(z) for z ∈ N_r^∗(z₀),

the result follows. 2

23.6 The principle of the argument 727 This gives another proof of the open mapping theorem.

Corollary 23.6.6 (The open mapping theorem) Suppose that f is a non-constant holomorphic function on a domain U . If V is an open subset of U then f (V ) is an open subset of C.

Proof Suppose that z₀ ∈ V . Let W be the connected component of V to which z₀ belongs, and apply the theorem to the restriction of f to W . If w ∈ Nρ(f (z₀)) then the equation f (z) = w has at least one solution in N_r(z₀), so that N_ρ(f (z₀))⊆ f(Nr(z₀))⊆ f(V ). Thus f(V ) is open. 2

It also provides another proof of the maximum modulus principle.

Corollary 23.6.7 Suppose that f is a non-constant holomorphic function on a domain U . Then |f| has no local maxima on U, and the only local minima are the zeros of f .

Proof If Nr(z0) is a neighbourhood of z0 contained in U , then f (z0) is an interior point of the open set f (Nr(z0)), and so|f(z0)| is not the supremum of|f| on Nr(z0), and is the infimum only if f (z0) = 0. 2

We now give an improved version of Theorem20.2.3.

Theorem 23.6.8 Suppose that f is a univalent function on a domain U . Then f (U ) is a domain, f is a homeomorphism of U onto f (U ), f^(z)= 0 for z∈ U, f⁻¹ : f (U )→ U is holomorphic and if f(z) = w then (f⁻¹)^(w) = 1/f^(z).

Proof If f^(z0) = 0 for some z0 ∈ U then the equation f(z) = w has more than one solution in U for values of w close to f (z₀), contradicting the fact that f is univalent. Thus f^(z) = 0 for z ∈ U. The derivative f^ is holomorphic, and is therefore continuous. The result is therefore follows

from Theorem20.2.3. 2

Exercises

23.6.1 Suppose that f is a non-constant continuous complex-valued func-tion on D whose restricfunc-tion to D is holomorphic. Show that if f (D)⊆ D then f has exactly one fixed point.

23.6.2 Suppose that|a| < 1. Show that the function z^m

 z− a 1− ¯az

_n

− a has m + n zeros in D.

23.6.3 Suppose that p(z) = a0+· · · + anzⁿ is a non-constant polynomial of degree n. Show that there exists R > 0 such that|p(z) − anzⁿ| <

|anzⁿ| for |z| ≥ R. Use Rouch´e’s theorem to give another proof of the fundamental theorem of algebra.

23.6.4 How many zeros does the function z sin z − 1 have in the disc N(n+1

2)π(0)? Use this to show that all the solutions of the equation z sin z = 1 are real.

23.6.5 (The inverse mapping theorem.) Suppose that f is a non-constant holomorphic function on a domain U and that z₀ ∈ U. What is the residue of the meromorphic function zf^(z)/(f (z)− f(z0)) at z0? Suppose that f is univalent, that γ is a contour in U and that V = in[γ]. If w∈ f(V ) let

g(w) = 1 2πi

γ

zf^(z) f (z)− wdz.

Show that g is the restriction of the inverse mapping f⁻¹ to f (V ).

23.6.6 Suppose that f is a meromorphic function on a domain U with the property that the residue at every pole is an integer. Suppose that z0 ∈ U \ Sf. If z∈ U \ Sf and γ is a rectifiable path in U\ Sf from z0 to z, let Fγ(z) = 

γf (z) dz. Show that e^Fγ(z) does not depend upon the choice of γ. Show that there exists a holomorphic function g on U\ Sf such that f = g^/g. Show further that g is meromorphic on U .

23.6.7 This exercise extends the results of Theorem 23.6.4.

(i) Suppose that U , f , z0 and d satisfy the conditions of Theorem 23.6.4 and that r and ρ satisfy its conclusions. Suppose that z₀ = 0 and that f (z₀) = 0. Show that there exists a holomorphic function h on U such that f (z) = z^dh(z) for z ∈ U, and that h(0)= 0.

(ii) Show that there exist 0 < r₁ < r and a univalent function k on N_r1(0) such that h(z) = k(z)^d for z∈ Nr1(0).

(iii) Let l(z) = zk(z) for z∈ Nr₁(0). Observe that f (z) = l(z)^d, for z ∈ Nr₁(0). Show that there exists 0 < r2 ≤ r1 such that l is univalent on Nr₂(0).

(iv) Let 0 < s < r2. Let γ0(t) = se^it for 0≤ t ≤ 2π/d. Let δ0 be the simple closed path σ(0, s)∨ γ0∨ σ(se^2πi/d, 0). Let 0 = l⁻¹◦ δ0, and let V₀ = in[₀]. Show that the restriction of f to V₀ is a univalent mapping of V0 onto the cut disc Ns(0)\ (−s, 0].

23.6 The principle of the argument 729 (v) Carry out similar constructions for the paths γj and δj, for 1≤ j < d, where γt(t) = se^it for 2πj/d≤ t ≤ 2π(j + 1)/d, and δ_j is the simple closed path σ(0, se^2πij/d)∨γj∨σ(se^2πi(j+1)/d, 0).

(vi) Draw a sketch to illustrate these constructions.

(vii) Show that there is no loss of generality in taking z₀ = 0 and f (z₀) = 0.

23.6.8 Let P_n = {a = (a0, . . . , a_n) ∈ Cⁿ⁺¹ : a_n = 0}. If a ∈ Pn, let r(a) = {z ∈ C : a0+ a₁z +· · · + anzⁿ = 0} be the set of roots of the polynomial pa(z) = a0+ a1z +· · · + anzⁿ, counted according to multiplicity. Explain why r(a) can be considered as an element of the weighted configuration space Wn(C) defined in the exercises of Volume II, Section 15.6.

Suppose that  > 0. Show that there exists a finite set Γ of disjoint circular paths in C\ r(a), each of radius less than , with centres the elements of r(a).

Let m = inf{|pa(z)| : z ∈ [Γ]}. Show that m > 0.

Show that there exists δ > 0 such that if b ∈ Pn and d(a, b) < δ then|pa(z)− pb(z)| < m for z ∈ [Γ].

Use Rouch´e’s theorem to show that the mapping r from P_n to (W_n(C), d_W) is continuous. (The roots of a polynomial depend continuously on the coefficients.)

23.6.9 Let Z(i) ={m + in : m, n ∈ Z} be the set of Gaussian integers. If z∈ C \ Z(i), let

f (z) = 1

z² + 

w∈Z(i)

 1

(z− w)² − 1 w²

 .

Prove carefully that the sum converges locally uniformly to a mero-morphic function f on C. Show that f (z + w) = f (z) for w∈ Z(i).

(f is doubly periodic.) Suppose that z0 ∈ C\Z(i) and that f(z0)= 0.

Show that f has two zeros (counted according to multiplicity) in the square with vertices z0, z0+ 1, z0+ 1 + i and z0+ i.

23.6.10 Suppose that f is a meromorphic function on C for which f (z) = f (z + w) for w ∈ Z(i). Show that if f is holomorphic, then f is constant. Suppose that f is not constant, and that f does not have a pole on the edges of the square with vertices z₀, z₀ + 1, z₀ + 1 + i and z₀ + i. Show that the sum of the residues of the poles within the square is zero. Show that the number of zeros within the square (counted according to multiplicity) is equal to the number of poles (counted according to multiplicity), and that the number is at least 2.