Power Series
3.5. Zeros and poles
with g having a power series expansion at ζ and g(ζ)= 0. We define N =νζ(f) = order (of the zero) of f at ζ .
Note that N ≥ 0, and N = 0 if and only iff(ζ) = 0. If N = 1, then we say that f has a simple zero atζ.
Notation. LetC =C∪ {∞} be the one point compactification of C, usually called the Riemann sphere. It is diffeomorphic to the unit sphere in R3. See Exercise 3.18.
Definition 3.44. (a) Let f be defined in a deleted neighborhood of ζ ∈C (see the Standard Notation summary). We say that
lim
z→ζf(z) =∞ if for all M >0, there exists a δ >0 such that
0<|z−ζ|< δ⇒ |f(z)|> M .
(b) Letα ∈C, and letf be defined in|z|> M for someM >0 (we say that f is defined in a deleted neighborhood of ∞ inC). We say
zlim→∞f(z) =α provided
zlim→0f 1
z =α.
(c) The above defines the concept of continuous maps between sets in C.
(d) A function f is holomorphic (has a power series expansion) at ∞ if and only if g(z) = f1
z
is holomorphic (has a power series expansion) at z = 0.
Definition 3.45. Let U ⊂ C be a neighborhood of a point ζ. A functionf that is holomorphic inU =U−{ζ}, a deleted neighborhood of a point ζ, has a removable singularity atζ if there is a holomorphic function in U that agrees with f onU.
Let us consider two functions f and g having power series expan- sions at each point of a domain Din C. Assume that neither function vanishes identically on D, and let ζ ∈ D∩C. Let F(z) = f(z)
(z−ζ)νζ(f)
3.5. ZEROS AND POLES 49
and G(z) = g(z)
(z−ζ)νζ(g) for z ∈D. Then the functions F and G have re- movable singularities at ζ, do not vanish there, and have power series expansions at each point of D. Furthermore
h(z) = f
g(z) = (z−ζ)νζ(f)F(z)
(z−ζ)νζ(g)G(z) for all ζ ∈D.
Exactly three distinct possibilities exist for the behavior of h(z) at ζ, which lead to the following definitions:
Definition 3.46. (I) νζ(g) > νζ(f). Then h(ζ) = ∞ [this defines h(ζ), and the resulting function h is continuous atζ]. In this case we say that h has a pole of order νζ(g)−νζ(f) at ζ. If νζ(g)−νζ(f) = 1, we say that the pole issimple.
(II) νζ(g) =νζ(f). The singularity of h atζ is removable, and by definition, h(ζ) = FG(ζ)(ζ).
(III) νζ(g)< νζ(f). The singularity is again removable and h(ζ) = 0.
In all cases we set νζ(h) = νζ(f)−νζ(g) and call it the order of h atζ.
In cases (II) and (III) of the definition, h has a power series expan- sion at ζ as a consequence of the following
Theorem 3.47. If f has a power series expansion atζ andf(ζ)= 0, then 1f also has a power series expansion at ζ.
Proof. Without loss of generality we assumeζ = 0 and f(0) = 1.
Thus
f(z) = ∞ n=0
anzn, a0 = 1, ρ >0 .
We want to find the reciprocal power series; that is, a power series g(z) = ∞
n=0bnzn with a positive radius of convergence and such that anzn bnzn
= 1.
The LHS and the RHS are both power series, where the RHS is a power series expansion whose coefficients for n > 1 are all zero.
Equating the first two coefficients, we obtain
a0b0 = 1, b0 = 1, a1b0+a0b1 = 0,
and using the n-th coefficient of the power series when expanded for the LHS, for n≥1, we obtain
anb0+an−1b1+. . .+a0bn= 0.
Thus by induction we define bn =−
n−1
j=0
bjan−j.
Since ρ > 0, we have 1ρ < +∞. Since lim supn|an|n1 = 1ρ, there exists a k > 0 such that |an| ≤kn.
We will show by induction that |bn| ≤ 2n−1kn for all n ≥ 1. For n = 1, we have b1 = −a1 and hence |b1| = |a1| ≤ k. Suppose the inequality holds for j ≤n for some n≥1. Then
|bn+1| ≤ n
j=0
|bj| |an+1−j|=|an+1|+ n
j=1
|bj| |an+1−j|
≤kn+1+ n j=1
2j−1kjkn+1−j =kn+1(1 + 2n−1).
Thus there is a reciprocal series, with radius of convergenceσ, satisfying 1
σ = lim sup
n |bn|n1 ≤lim
n (21−n1)k= 2k.
Corollary 3.48. LetD be a domain inC and f a function onD.
If f has a power series expansion at each point of D and f(z)= 0 for all z ∈D, then 1f has a power series expansion at each point of D.
Definition 3.49. For each domain D ⊆ C∪ {∞} =C, we define H(D) as
{f :D→C;f has a power series expansion at each point ofD}. The set H(D) is referred to as the set of holomorphic functions on D. We will see in Chapter 5 that this terminology is consistent with our earlier definition of a holomorphic function on D.
Corollary 3.50. Assume thatDis a domain inC. The setH(D) is an integral domain and an algebra over C. Its units are the functions that never vanish.
Definition 3.51. LetDbe a domain inC. A function f :D→C is meromorphic if it is locally4 the ratio of two functions having power series expansions (with the denominator not identically zero). The set of meromorphic functions on D is denoted by M(D).
4A propertyP is satisfied locally on an open set D if for each point a∈ D, there exists a neighborhoodU ⊂D ofasuch thatP is satisfied inU.
3.5. ZEROS AND POLES 51
Corollary 3.52. Let D be a domain in C, f ∈ M(D) and ζ ∈ D ∩C. Then there exist a connected neighborhood U of ζ in D, an integer n, and a unit g ∈H(U) such that
f(z) = (z−ζ)ng(z) for all z ∈U . Note that n =νζ(f).
Corollary 3.53. If D is a domain in C, then the set M(D) is a field and an algebra over C.
We recall that, by our convention, M(D)=0 is the set of meromor- phic functions with the constant function 0 omitted, where 0(z) = 0 for all z in D.
Corollary 3.54. If D be a domain and ζ ∈D, then νζ :M(D)=0 →Z
is a homomorphism.
Defining νζ(0) = +∞, we obtain
νζ(f +g)≥min{νζ(f), νζ(g)} for all f and g in M(D);
that is, νζ is a (discrete) valuation5 (of rank one) on M(D).
Remark 3.55. The converse statement also holds; it is non trivial and not established in this book.
The next corollary defines the term Laurent series.
Corollary 3.56. If f ∈ M(D)=0 and ζ ∈ D∩C, then f has a Laurent series expansion at ζ; that is, there exist aμ∈Z(μ=νζ(f)), a sequence of complex numbers {an}∞n=μ with aμ = 0, and a deleted neighborhood U of ζ such that
f(z) = ∞ n=μ
an(z−ζ)n for all z∈U. The power series
∞ n=max(0,μ)
an(z−ζ)n
converges uniformly and absolutely on compact subsets ofU =U∪{ζ}.
5Standard, but not universal, terminology.
Remark3.57.If∞ ∈D, then for all sufficiently large real numbers R the series representing f in {|z|> R} ∪ {∞} has the form
f(z) = ∞ n=μ
an
1 z
n
.
Corollary 3.58. If f ∈ M(D), then f ∈ M(D). If in addition νζ(f)= 0 for ζ ∈D, then
νζ(f) =νζ(f)−1 .
Remark 3.59. Some care must be exercised in discussing singular- ities and singular values. ∞is, of course, a singular value for holomor- phic functions but not for meromorphic ones.
Exercises
3.1. Determine the radius of convergence of each of the following series:
∞ n=0
zn n!,
∞ n=1
zn n ,
∞ n=0
n!zn .
3.2. Prove that if |an| ≤ M for n ≥ 0, then the power series ∞
n=0anzn has radius of convergence ρ≥1.
3.3.Under the hypothesis that{an}and{bn}are positive sequences, prove that:
(a)
limn anbn ≤lim
n anlim
n bn,
provided the right side is not the indeterminate form 0× ∞. Show by example that strict inequality may hold.
(b) If limnan exists, then the equality holds in (a) provided the right side is not indeterminate; that is, show that in this case,
limn anbn = lim
n anlim
n bn. 3.4. Give a proof of the ratio test.
3.5. Give a proof of Theorem 3.16.
3.6. Find all the roots of cosz = 2.
3.7. (a) Find (sinz),(sinz),(cosz),(cosz).
EXERCISES 53
(b) Write z =x+ıy and prove that
|sinz|2 = sin2x+ sinh2y and
|cosz|2 = cos2x+ sinh2y , where coshz = ez+e−z
2 and sinhz = ez−e−z
2 are the hyper- bolic trigonometric functions.
(c) Derive the addition formulas for cosh(a+b) and sinh(a+b).
(d) Evaluate Dsinhz, Dcoshz, and cosh2z−sinh2z.
3.8. Prove, using power series, thate−z = e1z.
3.9. Show that the exponential function is a transcendental func- tion.
3.10. Is it always true that Log(ez) =z? Support your answer with either a proof or a counterexample.
3.11. (a) What are all the possible values ofıı?
(b) Let a and b ∈ C with a = 0. Find necessary and sufficient conditions for ab to consist of infinitely many distinct values.
(c) Letn be a positive integer. Find necessary and sufficient con- ditions forab to consist of n distinct values.
3.12. (a) Show that both the sine and the cosine functions are pe- riodic with period 2π.
(b) Show that sinz = 0 if and only if z =πn for somen ∈Z. (c) Show that cosz = 0 if and only if z = π2(2n+ 1) for some
n∈Z.
3.13.Let{kn}be a sequence in which every positive integer appears once and only once.
Let
an be a series. Putting an = akn, we say that
an is a rearrangement of
an.
(a) Let {an} be a sequence of real numbers such that
an con- verges but
|an| does not. Let a be any real number. Show that there is a rearrangement
an of
an such that a = an.
(b) Show that
an converges absolutely if and only if every re- arrangement converges to the same sum.
3.14. Let{an} be a real sequence. Show that limn {an}= sup
α;α= lim
n bn
, with {bn} a convergent subsequence of {an} and that
lim
n
an= inf
α;α = lim
n bn with {bn} as above.
In this exercise a sequence {bn} with limbn = +∞ (similarly −∞) is to be considered a convergent sequence.
3.15. (Only for those who know some algebra.) Show that C is the only nontrivial finite-dimensional commutative division algebra overR. 3.16. Let p(z) = anzn +an−1zn−1 +...+a1z +a0, an = 0, be a polynomial of degree n≥1. Consider p as a self-map of C.
(a) Letα∈C. Show that there exists az ∈Csuch thatp(z) =α.6 (b) Let z ∈ C and p(z) = α ∈ C. Define appropriately mp(z), the multiplicity of α for pat z, so that you can prove: For all α∈C,
z∈C;p(z)=α
mp(z) =n. (3.10)
Note: The integer
z∈C;p(z)=α
mp(z) is thetopological degree of the map p:C →C.
3.17. Let p = PQ be a nonconstant rational map. It involves no loss of generality to assume, as we do, that P and Q do not have any common zeros. View, as in the case of polynomials, p as a self-map of C.
(a) Show that p is surjective.
(b) Define the concepts of multiplicity at a point and topological degree for the rational map p so that (3.10) holds.
3.18. The unit sphere (with center at 0)S2 ⊆R3 is defined by S2 ={(ξ, η, ζ)∈R3; ξ2+η2+ζ2 = 1}.
6You may use, although other arguments are available, the Fundamental The- orem of Algebra which will be established in Chapter 5.
EXERCISES 55
Show that stereographic projection
(ξ, η, ζ)→ ξ+ı η 1−ζ
is a diffeomorphism from S2− {(0,0,1)}ontoCand that it extends to a diffeomorphism from S2 ontoC (that sends (0,0,1) to ∞).
3.19. Justify the statement that stereographic projection takes cir- cles to circles.
That is, a “circle”onS2 is the intersection of a plane inR3 with S2. Such a circle is a “maximal circle” if it is the intersection of S2 with a plane through the point (0,0,1) of R3.
Show that stereographic projection sets up a bijective correspon- dence between the set of “maximal circles” onS2and the set of “circles”
through ∞ on C, that is, straight lines in C. Also show that stereo- graphic projection sets up a bijective correspondence between the set of all circles on S2 and what are called the circles in C: the union of the set of all circles in Cand the set of all straight lines in C.
The circles in C will play an important role in Chapter 8.
3.20. Show that stereographic projection preserves angles.
3.21. Suppose the power series ∞
−∞
αjzj and ∞
−∞
βjzj converge for 1<|z|<3 and 2<|z|<4, respectively, and that they have the same sum for 2<|z|<3. Does this imply that αj =βj for all j?
3.22. Find all zeroes of f(z) = 1−exp(expz).
3.23. Find the radius of convergence of the power series ∞
n=0
anzn,
where a0 = 0, a1 = 1, and an =an−1+an−2 for all n >1.
(Hint: Multiply the series by z2+z−1.) 3.24. The formula
tanz = sinz cosz
defines a meromorphic function on C. Show that it has simple poles at z = (2k+ 1)π
2 for every integer k
and is holomorphic elsewhere. Show that tan maps C ontoC∪ {∞}.
(a) Show that tanz = tanζ if and only if there exists an integer k such thatζ−z =πk.
(b) Show that z →tanz is a holomorphic one-to-one map of
z ∈C;−π
2 <z < π 2
onto C− {(−∞,−1)∪(1,+∞)} and of
z∈C;−π
2 <z ≤ π 2
ontoC∪ {∞}.7 (c) Show that
d
dz tanz = 1 cos2z.
3.25. One purpose of this exercise is to establish the beautiful for- mula (3.11).
Verify each of the following assertions and/or answer the questions:
(a) The series
1 1 +z2 =
∞ k=0
(−1)kz2k defines a holomorphic function on |z|<1.
(b) Hence there is a holomorphic function f(z) =
∞ k=0
(−1)kz2k+1 2k+ 1 on|z|<1 such that
f(0) = 0 and f(z) = 1 1 +z2.
(c) Since tan is locally injective, there exists a multivalued inverse function arctan defined onC∪ {∞} such that
tan(arctanz) =z for all z ∈C,
hence also tan(arctan(z) +kπ) =z for allk ∈Zand allz ∈C. We can then define theprincipal branch of arctan, to be called Arctan, by requiring that
−π
2 <(Arctan z)≤ π 2. (d) Show that
arctanz = 1
2ılog 1 +ız 1−ız.
7For this and the previous onto proof, you will need either some of the results of the next exercise or something like Rouch´e’s theorem, which is proven in Chapter 6.
EXERCISES 57
(e) Letg(z) =f(tanz). Show thatg(z) = 1 for allz in a domain D. Describe D.
(f) Conclude that f(z) = Arctan z for|z|<1.
(g) Why does the Taylor series for Arctan at the origin not con- verge in a bigger disk?
(h) Show that Arctan 1 is given by ∞
k=0 (−1)k
2k+1, thus justifying π = 4
∞ k=0
(−1)k
2k+ 1. (3.11)
3.26. (l’Hopital’s rule) Let f and g be two functions defined by convergent power series in a neighborhood of 0. Assume that f(0) = 0 = g(0) and g(0)= 0. Show that
zlim→0
f(z)
g(z) = f(0) g(0).