Zeros of Analytic Functions

Lecture 26

(ii). In view of Problem 24.4, the function g(z) defined in (18.7) has no zero at 0.

(iii). Since

f₁(z) = z²sinz = z²

z−z³ 3! +z⁵

5! − · · ·

= z³

1−z² 3! +z⁴

5! − · · ·

, f₁(z) has a zero of order 3 atz= 0.

(iv). Since

f2(z) = zsinz² = z

z²−z⁶ 3! +z¹⁰

5! − · · ·

= z³

1−z⁴ 3! +z⁸

5! − · · ·

, f₂(z) has a zero of order 3 atz= 0.

Corollary 26.2.

Iff(z) is analytic at z0 and f(z0) = 0, then either f(z) is identically zero in a neighborhood ofz₀or there is a punctured disk aboutz₀ in whichf(z) has no zeros.

Proof.

Let $_∞

j=0aj(z−z0)^j be the Taylor series for f(z) about z0. This series converges to f(z) in some neighborhood of z₀. So, if all the Taylor coefficients aj are zero, then f(z) must be identically zero in this neighborhood. Otherwise, let m ≥ 1 be the smallest subscript such that am= 0.Then, f(z) has a zero of orderm atz₀,and so the representation f(z) = (z−z₀)^mg(z) by Theorem 26.1 is valid. Sinceg(z₀)= 0 andg(z) is continuous at z₀, there exists a disk B(z₀, δ) throughout which g(z) is nonzero. Consequently,f(z)= 0 for 0<|z−z₀|< δ.

Thus, for a functionf(z) that is analytic and has a zero at a point z0

but is not identically equal to zero in any neighborhood ofz0,the pointz0

must be a zero of some finite order; and conversely, ifz0 is a zero of finite order of f(z), then f(z) = 0 throughout some punctured neighborhood N(z0) = {z : 0 <|z−z0| < } of z0. Thus, a finite-order zero off(z) is isolated from other zeros off(z).

Corollary 26.3.

Supposef(z) is analytic on a regionS,and{z_n}is an infinite sequence of distinct points in S converging to α∈S. If f(zn) = 0 for alln,thenf(z) is identically zero on S.

Proof.

Sincef(z) is continuous, 0 = lim_n→∞f(z_n) = f(α). We claim thatf(z) is identically zero in some neighborhood ofα.In fact, iff(z)= 0 in some punctured neighborhood N(α), then, from the definition of the

Zeros of Analytic Functions 179 limit, for sufficiently largen,there existsz_n ∈N(α) such thatf(z_n) = 0.

The rest of the proof is exactly the same as that of Theorem 20.1 (recall Figure 20.1).

Corollary 26.4.

If a function f(z) is analytic on a domain S and f(z) = 0 at each pointzof a subdomain ofSor an arc contained inS,then f(z)≡0 inS.

Now supposef(z) andg(z) are analytic on the same domainSand that f(z) =g(z) at each pointz of a subdomain ofS or an arc contained inS.

Then, the function h(z) defined byh(z) =f(z)−g(z) is also analytic on S and h(z) = 0 throughout the subdomain or along the arc. But then, in view of Corollary 26.4, h(z) = 0 throughout S; i.e., f(z) = g(z), z ∈ S.

This proves the following corollary.

Corollary 26.5.

A function that is analytic on a domainSis uniquely determined over S by its values over a subdomain of S or along an arc contained inS.

Example 26.2.

(i). Since sin²x+ cos²x= 1, the entire function f(z) = sin²z+cos²z−1 has zero values along the real axis. Thus, Corollary 26.4 implies that f(z) = 0 throughout the complex plane; i.e., sin²z+ cos²z= 1 for allz.Furthermore, Corollary 26.5 tells us that sinzand cosz are the only entire functions that can assume the values sinx and cosx, respectively, along the real axis or any segment of it.

(ii). We can use cosh²x−sinh²x= 1 to show that cosh²z−sinh²z= 1 for all complexz.

(iii). We can use sin(x1+x2) = sinx1cosx2+ cosx1sinx2 to show that for all complex z1, sin(z1+x2) = sinz1cosx2 + cosz1sinx2, and from this deduce that sin(z1+z2) = sinz1cosz2+ cosz1sinz2 for all complex numbersz1 andz2.

(iv). We can use e^x1+x2 = e^x1e^x2 to show that e^z1+z2 = e^z1e^z2 for all complex numbersz₁and z₂.

Corollary 26.6.

If a functionf(z) is analytic on a bounded domain S and continuous and nonvanishing on the boundary of S, then f(z) can have at most finitely many zeros insideS.

Proof.

Suppose that f(z) has an infinite number of zeros inside S.

SinceS is closed and bounded, in view of Theorem 4.2, there is an infinite sequence of zeros{z_n} that converges to a pointα∈S.Now, sincef(z) is continuous onS, f(α) = lim_n→∞f(z_n) = 0,and sincef(z) is nonvanishing on the boundary,αmust be inside S.But then, by Corollary 26.3, f(z) is identically zero on S,and since f(z) is continuous it must be zero on the boundary ofS also. This contradiction completes the proof.

From Corollary 26.6, it follows that an analytic function can have an infinite number of zeros only in an open or unbounded domain. From Corollary 26.6, it also follows that an entire function in any bounded part of the complex plane can have only a finite number of zeros. Thus, all the zeros of an entire function can be arranged in some kind of order, for example in order of increasing absolute value. In the extended complex plane, an entire function can have only a countable set of zeros, and the limit pointof this set is the point at infinity of the complex plane.

Extension of known equalities from real to complex variables as in Ex- ample 26.2 can be generalized to a broader class of identities. In the follow- ing result we state such a generalization for an important class of identities that involve only polynomials of functions.

Theorem 26.2.

LetP(f1,· · ·, fn) be a polynomial in the nvariables fj, j= 1,· · ·, n,where eachfj is an analytic function ofz in a domainS that contains some intervala < x < bof the real axis. If for allx∈(a, b)

P(f₁(x),· · ·, fn(x)) = 0, then for allz∈S

P(f1(z),· · ·, fn(z)) = 0.

Proof.

Clearly, the function P(f₁(z),· · ·, fn(z)) is analytic in S and vanishes over an arc inS.The result now follows from Corollary 26.4.

We conclude this lecture by proving the following result.

Theorem 26.3 (Counting Zeros).

Let f(z) be analytic inside and on a positively oriented contour γ. Furthermore, let f(z) = 0 on γ.

Then,

1 2πi

γ

f(z)

f(z)dz = Z_f (26.1)

holds, where Zf is the number of zeros (counting with multiplicities) of f(z) that lie insideγ.

Proof.

The functionf(z)/f(z) is analytic inside and onγexcept at the zerosa₁,· · ·, aoff(z) lying insideγ.Supposem₁,· · ·, mare the respective multiplicities of these zeros. ThenZ_f =m₁+· · ·+m.In view of Theorem 26.1, we can find disjoint open disks B(a_k, r_k), k= 1,· · ·, , and analytic nonzero functionsg_k(z) inB(a_k, r_k) such that

f(z) = (z−ak)^mkgk(z), z∈B(ak, rk). (26.2) Thus, we have

f(z)

f(z) = mk

z−ak

+g_k(z)

gk(z), z∈B(ak, rk)\{ak}. (26.3)

Zeros of Analytic Functions 181 Consider the function

F(z) =

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩ f(z)

f(z) − j=1

mj

z−a_j, z∈ ) j=1

B(aj, rj), g_k(z)

g_k(z)− j=1,j=k

mj

z−a_j, z∈B(ak, rk), k= 1,· · ·, . Clearly,F(z) is analytic inside and onγ,and hence

γ

F(z)dz= 0.There- fore, from Theorem 16.2 and Example 16.1, it follows that

γ

f(z)

f(z)dz = 2πi j=1

mj = 2πiZf, which is the same as (26.1).

Problems

26.1. Locate and determine the order of zeros of the following functions:

(a). e^2z−e^z, (b). z²sinhz, (c). z⁴cos²z, (d). z³cosz².

26.2. Suppose that f(z) is analytic and has a zero of orderm at z0. Show that

(a). f(z) has a zero of orderm−1 atz₀. (b). f(z)/f(z) has a simple pole atz₀.

26.3.Find an entire functionf(z) with prescribed distinct zerosz1,· · ·, zk

with multiplicitiesm₁,· · ·, m_k,respectively. Isf(z) uniquely determined?

26.4. Iff(z) is analytic on a domainSand has distinct zerosz1,· · ·, zk

with multiplicitiesm1,· · ·, mk,respectively, show that there exists an ana- lytic functiong(z) on S such thatf(z) = (z−z₁)^m1· · ·(z−z_k)^mkg(z).

26.5.Suppose thatf(z) andg(z) are analytic in a regionSandf(z)g(z) is identically zero inS.Show that eitherf(z) org(z) is identically zero in S.

Answers or Hints

26.1. (a). z = 2kπi, k = 0,±1,±2,· · · simple zeros, (b). z = 0 zero of order 3, z =kiπ, k=±1,±2,· · · simple zeros, (c). z= 0 zero of order 4, z = (2k+ 1)π/2, k= 0,±1,±2,· · · simple zeros, (d).z= 0 zero of order 3, z=±

(2k+ 1)π/2, ±i

(2k+ 1)π/2, k= 0,1,2,· · ·simple zeros.

26.2. By Theorem 26.1,f(z) = (z−z₀)^mg(z),wheregis analytic atz₀and g(z₀)= 0. (a). f(z) = (z−z₀)^m−1h(z),whereh(z) =mg(z)+(z−z₀)g(z) is analytic atz₀andh(z₀) =mg(z₀)= 0,sof has a zero of orderm−1 at z₀by Theorem 26.1 again. (b). f(z)/f(z) =p(z)/(z−z₀),wherep=h/g is analytic atz₀ andp(z₀) =m= 0,so f/f has a simple pole atz₀. 26.3. We can take, for example,f(z) =*k

j=1(z−z_j)^mj; it is not uniquely determined since multiplying it by an entire function without zeros gives another such function.

26.4. Taylor’s expansion of f(z) at z₁ yields an analytic function f₁(z) such that f(z) = (z−z₁)^m1f₁(z), where z₁ is not a zero of f₁(z). Now expandf₁(z) atz₂.

26.5. If neitherf norgis identically zero inS,then they both have isolated zeros, so the zeros ofh=f gare also isolated, and hencehis not identically zero inS.

Lecture 27