Capítulo 5 Estrategia de Solución
5.1 Identificación de las variables, señales o estados de interés
Let f have an isolated essential singularity at z0. Then for any complex number w,
f(z)comes arbitrarily close to w in every deleted neighborhood ofz0. That is, for any
δ >0, f(D′(z
0, δ)) is a dense subset ofC.
Proof. Suppose that for someδ >0, f(D′(z
0, δ)) is not dense inC. Then for somew∈C,
there exists ǫ > 0 such that D(w, ǫ)does not meet f(D′(z
g(z) = 1/(f(z)−w)). Thengis bounded and analytic onD′(z
0, δ), and hence by (4.1.5a),
ghas a removable singularity atz0. Letmbe the order of the zero ofgatz0(setm= 0 if
g(z0)= 0)and writeg(z) = (z−z0)mg1(z)whereg1is analytic onD(z0, δ)andg1(z0)= 0
[see (2.4.4)]. Then (z−z0)mg1(z) = 1/(f(z)−w) , so asz approachesz0,
(z−z0)mf(z) = (z−z0)mw+ 1 g1(z) −→ w+ 1/g1(z0)ifm= 0 1/g1(z0)ifm= 0.
Thusf has a removable singularity or a pole atz0. ♣
4.1.7
Remark
The Casorati-Weierstrass theorem is actually a weak version of a much deeper result called the “big Picard theorem”, which asserts that iff has an isolated essential singularity at
z0, then for anyδ >0, f(D′(z0, δ)) is either the complex plane Cor Cminus one point.
We will not prove this result.
The behavior of a complex function f at ∞ may be studied by considering g(z) =
f(1/z)forznear 0. This allows us to talk about isolated singularities at∞. Here are the formal statements.
4.1.8
Definition
We say thatf has anisolated singularity at∞iff is analytic on{z:|z|> r} for somer; thus the functiong(z) =f(1/z)has an isolated singularity at 0. The type of singularity off at∞is then defined as that ofg at 0.
4.1.9
Remark
Liouville’s theorem implies that if an entire functionf has a removable singularity at∞, thenf is constant. (By (4.1.5a),f is bounded onC.)
Problems
1. Complete the proofs of (a)and (b)of (4.1.5). (Hint for (a): If f is bounded on
D′(z
0, δ), considerg(z) = (z−z0)f(z).)
2. Classify the singularities of each of the following functions (include the point at ∞). (a) z/sinz (b)exp(1/z)(c)zcos 1/z (d)1/[z(ez
−1)] (e) cotz
3. Obtain three different Laurent expansions of (7z−2)/z(z+ 1) (z−2)aboutz=−1. (Use partial fractions.)
4. Obtain all Laurent expansions of f(z) = z−1+ (z
−1)−2+ (z+ 2)−1 about z = 0,
and indicate where each is valid.
5. Find the first few terms in the Laurent expansion of z2(ez1−e−z) valid for 0<|z|< π.
6. Without carrying out the computation in detail, indicate a relatively easy procedure for finding the Laurent expansion of 1/sinz valid forπ <|z|<2π.
6 CHAPTER 4. APPLICATIONS OF THE CAUCHY THEORY
7. (Partial Fraction Expansion). LetR(z) =P(z)/Q(z), whereP andQare polynomials and degP < degQ. (If this is not the case, then by long division we may write
P(z)/Q(z) =anzn+· · ·+a1z+a0+P1(z)/Q(z)where degP1 <degQ.)Suppose
that the zeros of Q are at z1, . . . , zk with respective orders n1, . . . , nk. Show that R(z) =k
j=1Bj(z), whereBj(z)is of the form Aj,0 (z−zj)nj +· · ·+Aj,(nj−1) (z−zj) , with Aj,r= lim z→zj 1 r! dr dzr[(z−zj) njR(z)]
(dzdrrf(z)is interpreted asf(z)whenr= 0) .
Apply this result toR(z) = 1/[z(z+i)3].
8. Find the sum of the series ∞
n=0e−nsinnz (in closed form), and indicate where
the series converges. Make an appropriate statement about uniform convergence. (Suggestion: Consider∞
n=0e−neinz and
∞
n=0e−ne−inz.)
9. (a)Show that iff is analytic on ˆC, thenf is constant.
(b)Suppose f is entire and there exists M >0 and k >0 such that |f(z)| ≤M|z|k for |z| sufficiently large. Show thatf(z)is a polynomial of degree at mostk. (This can also be done without series; see Problem 2.2.13.)
(c)Prove that if f is entire and has a nonessential singularity at ∞, then f is a polynomial.
(d)Prove that iff is meromorphic on ˆC(that is, any singularity off in ˆCis a pole), thenf is a rational function.
10. Classify the singularities of the following functions (include the point at∞). (a)sin 2z z4 (b) 1 z2(z+ 1)+ sin 1 z (c)cscz− k z (d)exp(tan 1 z)(e) 1 sin(sinz). 11. Suppose thataandbare distinct complex numbers. Show that (z−a)/(z−b)has an
analytic logarithm onC\[a, b], call itg. Then find the possible Laurent expansions ofg(z)aboutz= 0.
12. Supposef is entire andf(C)is not dense inC. Show thatf is constant.
13. Assumef has a pole of ordermatα, andP is a polynomial of degreen. Prove that the composition P◦f has a pole of ordermn atα.
4.2
Residue Theory
We now develop a technique that often allows for the rapid evaluation of integrals of the form
γf(z)dz, where γ is a closed path (or cycle)in Ω and f is analytic on Ω except possibly for isolated singularities.
4.2.1
Definition
Letf have an isolated singularity at z0, and let the Laurent expansion off aboutz0 be
∞
n=−∞an(z−z0)n. Theresidue off atz0, denoted by Res(f, z0), is defined to bea−1.
4.2.2
Remarks
In many cases, the evaluation of an integral can be accomplished by the computation of residues. This is illustrated by (a)and (b)below.
(a)Supposef has an isolated singularity atz0, so thatf is analytic onD′(z0, ρ)for some
ρ >0. Then for anyrsuch that 0< r < ρ, we have
C(z0,r)
f(w)dw= 2πiRes(f, z0).
Proof. Apply the integral formula (4.1.3)fora−1. ♣
(b)More generally, if γ is a closed path or cycle inD′(z
0, ρ)such thatn(γ, z0)= 1 and
n(γ, z)= 0 for everyz /∈D(z0, ρ), then
γ
f(w)dw= 2πiRes(f, z0).
Proof. This follows from (3.3.7). ♣
(c)Res(f, z0)is that numberksuch thatf(z)−[k/(z−z0)] has a primitive onD′(z0, ρ).
Proof. Note that if 0< r < ρ, then by (a), C(z0,r) f(w)−w k −z0 dw= 2πi[Res(f, z0)−k].
Thus iff(z)−[k/(z−z0)] has a primitive onD′(z0, ρ), then the integral is zero, and hence
Res(f, z0) =k. Conversely, if Res(f, z0) =k, then
f(z)− k z−z0 = ∞ n=−∞ n=−1 an(z−z0)n,
which has a primitive onD′(z
0, ρ), namely ∞ n=−∞ n=−1 an n+ 1(z−z0) n+1. ♣
(d)Iff has a pole of ordermatz0, then
Res(f, z0) = 1 (m−1)!zlim→z0 dm−1 dzm−1[(z−z0) mf(z)].
8 CHAPTER 4. APPLICATIONS OF THE CAUCHY THEORY
In particular, iff has a simple pole atz0, then
Res(f, z0)= lim
z→z0
[(z−z0)f(z)].
Proof. Let {an} be the Laurent coefficient sequence for f about z0, so that an = 0 for n <−manda−m= 0. Then forz∈D′(z0, ρ),
(z−z0)mf(z) =a−m+a−m+1(z−z0) +· · ·+a−1(z−z0)m−1+a0(z−z0)m+· · · , hence dm−1 dzm−1[(z−z0) mf(z) ] = (m −1)!a−1+ (z−z0)g(z)
whereg has a removable singularity atz0. The result follows. ♣
(e)Supposef is analytic atz0 and has a zero of orderk at z0. Then f′/f has a simple
pole atz0 and Res(f′/f, z0) =k.
Proof. There exists ρ > 0 and a zero-free analytic function g on D(z0, ρ)such that
f(z) = (z−z0)kg(z)for z ∈ D(z0, ρ). Then f′(z) =k(z−z0)k−1g(z) + (z−z0)kg′(z),
and hence forz∈D′(z
0, ρ), f′(z) f(z) = k z−z0 + g′(z) g(z).
Since g′/g is analytic on D(z0, ρ), it follows that f′/f has a simple pole at z0 and
Res(f′/f, z
0) =k. ♣
We are now ready for the main result of this section.
4.2.3
Residue Theorem
Letf be analytic on Ω\S, where S is a subset of Ω with no limit point in Ω. In other words,f is analytic on Ω except for isolated singularities. Then for any closed path (or cycle)γ in Ω\S such thatγ is Ω-homologous to 0, we have
γ
f(w)dw= 2πi w∈S
n(γ, w)Res(f, w).
Proof. Let S1 = {w ∈ S : n(γ, w) = 0}. Then S1 ⊆Q = C\ {z /∈ γ∗ : n(γ, z) = 0}.
Sinceγis Ω-homologous to 0,Qis a subset of Ω. Furthermore, by (3.2.5),Qis closed and bounded. SinceShas no limit point in Ω,S1has no limit points at all. ThusS1is a finite
set. Consequently, the sum that appears in the conclusion of the theorem is the finite sum obtained by summing overS1. Let w1, w2, . . . , wk denote the distinct points ofS1.
[IfS1 is empty, we are finished by Cauchy’s theorem (3.3.1).] Choose positive numbers
r1, r2, . . . , rk so small that
D′(wj, rj)⊆Ω\S, j= 1,2, . . . , k. Let σ be the cycle k
j=1n(γ, wj)γj, where γj is the positively oriented boundary of
n(γ, z) =n(σ, z). Sincef is analytic on Ω\S, it follows from (3.3.7)that γf(w)dw = σf(w)dw. But by definition ofσ, γ f(w)dw= k j=1 n(γ, wj) γj f(w)dw= 2πi k j=1 n(γ, wj)Res(f, wj) by part (a)of (4.2.2). ♣
In many applications of the residue theorem, the integral
γf(w)dw is computed by evaluating the sum 2πi
w∈Sn(γ, w)Res(f, w). Thus it is important to have methods available for calculating residues. For example, (4.2.2d)is useful when f is a rational function, since the only singularities of f are poles. The residue theorem can also be applied to obtain a basic geometric property of analytic functions called the argument principle. Before discussing the general result, let’s look at a simple special case. Suppose
z traverses the unit circle once in the positive sense, that is, z =eit,0 ≤t≤2π. Then the argument of z2, namely 2t, changes by 4π, so thatz2 makes two revolutions around
the origin. Thus the number of times thatz2 winds around the origin asztraverses the
unit circle is the number of zeros ofz2 inside the circle, counting multiplicity.
The index of a point with respect to a closed path allows us to formalize the notion of the number of times thatf(z)winds around the origin asztraverses a pathγ. For we are looking at the net number of revolutions about 0 off(γ(t)), a≤t≤b, and this, as we have seen, isn(f◦γ,0). We may now state the general result.