Digital Library of Mathematical Functions
About the Project
NIST
1 Algebraic and Analytic MethodsAreas

§1.10 Functions of a Complex Variable

Contents

§1.10(i) Taylor’s Theorem for Complex Variables

Let f(z) be analytic on the disk |z-z0|<R. Then

1.10.1 f(z)=n=0f(n)(z0)n!(z-z0)n.

The right-hand side is the Taylor series for f(z) at z=z0, and its radius of convergence is at least R.

Examples

1.10.2 z=1+z1!+z22!+,
|z|<,
1.10.3 ln(1+z)=z-z22+z33-,
|z|<1,
1.10.4 (1-z)-α=1+αz+α(α+1)2!z2+α(α+1)(α+2)3!z3+,
|z|<1.

Again, in these examples ln(1+z) and (1-z)-α have their principal values; see §§4.2(i) and 4.2(iv).

Zeros

An analytic function f(z) has a zero of order (or multiplicity) m (1) at z0 if the first nonzero coefficient in its Taylor series at z0 is that of (z-z0)m. When m=1 the zero is simple.

§1.10(ii) Analytic Continuation

Let f1(z) be analytic in a domain D1. If f2(z), analytic in D2, equals f1(z) on an arc in D=D1D2, or on just an infinite number of points with a limit point in D, then they are equal throughout D and f2(z) is called an analytic continuation of f1(z). We write (f1,D1), (f2,D2) to signify this continuation.

Suppose z(t)=x(t)+y(t), atb, is an arc and a=t0<t1<<tn=b. Suppose the subarc z(t), t[tj-1,tj] is contained in a domain Dj, j=1,,n. The function f1(z) on D1 is said to be analytically continued along the path z(t), atb, if there is a chain (f1,D1), (f2,D2),,(fn,Dn).

Analytic continuation is a powerful aid in establishing transformations or functional equations for complex variables, because it enables the problem to be reduced to: (a) deriving the transformation (or functional equation) with real variables; followed by (b) finding the domain on which the transformed function is analytic.

Schwarz Reflection Principle

Let C be a simple closed contour consisting of a segment AB of the real axis and a contour in the upper half-plane joining the ends of AB. Also, let f(z) be analytic within C, continuous within and on C, and real on AB. Then f(z) can be continued analytically across AB by reflection, that is,

1.10.5 f(z¯)=f(z)¯.

§1.10(iii) Laurent Series

Suppose f(z) is analytic in the annulus r1<|z-z0|<r2, 0r1<r2, and r(r1,r2). Then

1.10.6 f(z)=n=-an(z-z0)n,

where

1.10.7 an=12π|z-z0|=rf(z)(z-z0)n+1z,

and the integration contour is described once in the positive sense. The series (1.10.6) converges uniformly and absolutely on compact sets in the annulus.

Let r1=0, so that the annulus becomes the punctured neighborhood N: 0<|z-z0|<r2, and assume that f(z) is analytic in N, but not at z0. Then z=z0 is an isolated singularity of f(z). This singularity is removable if an=0 for all n<0, and in this case the Laurent series becomes the Taylor series. Next, z0 is a pole if an0 for at least one, but only finitely many, negative n. If -n is the first negative integer (counting from -) with a-n0, then z0 is a pole of order (or multiplicity) n. Lastly, if an0 for infinitely many negative n, then z0 is an isolated essential singularity.

The singularities of f(z) at infinity are classified in the same way as the singularities of f(1/z) at z=0.

An isolated singularity z0 is always removable when limzz0f(z) exists, for example (sinz)/z at z=0.

The coefficient a-1 of (z-z0)-1 in the Laurent series for f(z) is called the residue of f(z) at z0, and denoted by resz=z0[f(z)], resz=z0[f(z)], or (when there is no ambiguity) res[f(z)].

A function whose only singularities, other than the point at infinity, are poles is called a meromorphic function. If the poles are infinite in number, then the point at infinity is called an essential singularity: it is the limit point of the poles.

Picard’s Theorem

In any neighborhood of an isolated essential singularity, however small, an analytic function assumes every value in with at most one exception.

§1.10(iv) Residue Theorem

If f(z) is analytic within a simple closed contour C, and continuous within and on C—except in both instances for a finite number of singularities within C—then

1.10.8 12πCf(z)z=sum of the residues of f(z) within C.

Here and elsewhere in this subsection the path C is described in the positive sense.

Phase (or Argument) Principle

If the singularities within C are poles and f(z) is analytic and nonvanishing on C, then

1.10.9 N-P=12πCf(z)f(z)z=12πΔC(phf(z)),

where N and P are respectively the numbers of zeros and poles, counting multiplicity, of f within C, and ΔC(phf(z)) is the change in any continuous branch of ph(f(z)) as z passes once around C in the positive sense. For examples of applications see Olver (1997b, pp. 252–254).

In addition,

1.10.10 12πCzf(z)f(z)z=(sum of locations of zeros)-(sum of locations of poles),

each location again being counted with multiplicity equal to that of the corresponding zero or pole.

Rouché’s Theorem

If f(z) and g(z) are analytic on and inside a simple closed contour C, and |g(z)|<|f(z)| on C, then f(z) and f(z)+g(z) have the same number of zeros inside C.

§1.10(v) Maximum-Modulus Principle

Analytic Functions

If f(z) is analytic in a domain D, z0D and |f(z)||f(z0)| for all zD, then f(z) is a constant in D.

Let D be a bounded domain with boundary D and let D¯=DD. If f(z) is continuous on D¯ and analytic in D, then |f(z)| attains its maximum on D.

Harmonic Functions

If u(z) is harmonic in D, z0D, and u(z)u(z0) for all zD, then u(z) is constant in D. Moreover, if D is bounded and u(z) is continuous on D¯ and harmonic in D, then u(z) is maximum at some point on D.

Schwarz’s Lemma

In |z|<R, if f(z) is analytic, |f(z)|M, and f(0)=0, then

1.10.11 |f(z)|M|z|R and |f(0)|MR.

Equalities hold iff f(z)=Az, where A is a constant such that |A|=M/R.

§1.10(vi) Multivalued Functions

Functions which have more than one value at a given point z are called multivalued (or many-valued) functions. Let F(z) be a multivalued function and D be a domain. If we can assign a unique value f(z) to F(z) at each point of D, and f(z) is analytic on D, then f(z) is a branch of F(z).

Example

F(z)=z is two-valued for z0. If D=\(-,0] and z=rθ, then one branch is rθ/2, the other branch is -rθ/2, with -π<θ<π in both cases. Similarly if D=\[0,), then one branch is rθ/2, the other branch is -rθ/2, with 0<θ<2π in both cases.

A cut domain is one from which the points on finitely many nonintersecting simple contours (§1.9(iii)) have been removed. Each contour is called a cut. A cut neighborhood is formed by deleting a ray emanating from the center. (Or more generally, a simple contour that starts at the center and terminates on the boundary.)

Suppose F(z) is multivalued and a is a point such that there exists a branch of F(z) in a cut neighborhood of a, but there does not exist a branch of F(z) in any punctured neighborhood of a. Then a is a branch point of F(z). For example, z=0 is a branch point of z.

Branches can be constructed in two ways:

(a) By introducing appropriate cuts from the branch points and restricting F(z) to be single-valued in the cut plane (or domain).

(b) By specifying the value of F(z) at a point z0 (not a branch point), and requiring F(z) to be continuous on any path that begins at z0 and does not pass through any branch points or other singularities of F(z).

If the path circles a branch point at z=a, k times in the positive sense, and returns to z0 without encircling any other branch point, then its value is denoted conventionally as F((z0-a)2kπ+a).

Example

Let α and β be real or complex numbers that are not integers. The function F(z)=(1-z)α(1+z)β is many-valued with branch points at ±1. Branches of F(z) can be defined, for example, in the cut plane D obtained from by removing the real axis from 1 to and from -1 to -; see Figure 1.10.1. One such branch is obtained by assigning (1-z)α and (1+z)β their principal values (§4.2(iv)).

See accompanying text
Figure 1.10.1: Domain D. Magnify

Alternatively, take z0 to be any point in D and set F(z0)=αln(1-z0)βln(1+z0) where the logarithms assume their principal values. (Thus if z0 is in the interval (-1,1), then the logarithms are real.) Then the value of F(z) at any other point is obtained by analytic continuation.

Thus if F(z) is continued along a path that circles z=1 m times in the positive sense and returns to z0 without circling z=-1, then F((z0-1)2mπ+1)=αln(1-z0)βln(1+z0)2πmα. If the path also circles z=-1 n times in the clockwise or negative sense before returning to z0, then the value of F(z0) becomes αln(1-z0)βln(1+z0)2πmα-2πnβ.

§1.10(vii) Inverse Functions

Lagrange Inversion Theorem

Suppose f(z) is analytic at z=z0, f(z0)0, and f(z0)=w0. Then the equation

1.10.12 f(z)=w

has a unique solution z=F(w) analytic at w=w0, and

1.10.13 F(w)=z0+n=1Fn(w-w0)n

in a neighborhood of w0, where nFn is the residue of 1/(f(z)-f(z0))n at z=z0. (In other words nFn is the coefficient of (z-z0)-1 in the Laurent expansion of 1/(f(z)-f(z0))n in powers of (z-z0); compare §1.10(iii).)

Furthermore, if g(z) is analytic at z0, then

1.10.14 g(F(w))=g(z0)+n=1Gn(w-w0)n,

where nGn is the residue of g(z)/(f(z)-f(z0))n at z=z0.

Extended Inversion Theorem

Suppose that

1.10.15 f(z)=f(z0)+n=0fn(z-z0)μ+n,

where μ>0, f00, and the series converges in a neighborhood of z0. (For example, when μ is an integer f(z)-f(z0) has a zero of order μ at z0.) Let w0=f(z0). Then (1.10.12) has a solution z=F(w), where

1.10.16 F(w)=z0+n=1Fn(w-w0)n/μ

in a neighborhood of w0, nFn being the residue of 1/(f(z)-f(z0))n/μ at z=z0.

It should be noted that different branches of (w-w0)1/μ used in forming (w-w0)n/μ in (1.10.16) give rise to different solutions of (1.10.12). Also, if in addition g(z) is analytic at z0, then

1.10.17 g(F(w))=g(z0)+n=1Gn(w-w0)n/μ,

where nGn is the residue of g(z)/(f(z)-f(z0))n/μ at z=z0.

§1.10(viii) Functions Defined by Contour Integrals

Let D be a domain and [a,b] be a closed finite segment of the real axis. Assume that for each t[a,b], f(z,t) is an analytic function of z in D, and also that f(z,t) is a continuous function of both variables. Then

1.10.18 F(z)=abf(z,t)t

is analytic in D and its derivatives of all orders can be found by differentiating under the sign of integration.

This result is also true when b=, or when f(z,t) has a singularity at t=b, with the following conditions. For each t[a,b), f(z,t) is analytic in D; f(z,t) is a continuous function of both variables when zD and t[a,b); the integral (1.10.18) converges at b, and this convergence is uniform with respect to z in every compact subset S of D.

The last condition means that given ϵ (>0) there exists a number a0[a,b) that is independent of z and is such that

1.10.19 |a1bf(z,t)t|<ϵ,

for all a1[a0,b) and all zS; compare §1.5(iv).

M-test

If |f(z,t)|M(t) for zS and abM(t)t converges, then the integral (1.10.18) converges uniformly and absolutely in S.

§1.10(ix) Infinite Products

Let pk,m=n=km(1+an). If for some k1, pk,mpk0 as m, then we say that the infinite product n=1(1+an) converges. (The integer k may be greater than one to allow for a finite number of zero factors.) The convergence of the product is absolute if n=1(1+|an|) converges. The product n=1(1+an), with an-1 for all n, converges iff n=1ln(1+an) converges; and it converges absolutely iff n=1|an| converges.

Suppose an=an(z), zD, a domain. The convergence of the infinite product is uniform if the sequence of partial products converges uniformly.

M-test

Suppose that an(z) are analytic functions in D. If there is an N, independent of zD, such that

1.10.20 |ln(1+an(z))|Mn,
nN,

and

1.10.21 n=1Mn<,

then the product n=1(1+an(z)) converges uniformly to an analytic function p(z) in D, and p(z)=0 only when at least one of the factors 1+an(z) is zero in D. This conclusion remains true if, in place of (1.10.20), |an(z)|Mn for all n, and again n=1Mn<.

Weierstrass Product

If {zn} is a sequence such that n=1|zn-2| is convergent, then

1.10.22 P(z)=n=1(1-zzn)z/zn

is an entire function with zeros at zn.

§1.10(x) Infinite Partial Fractions

Suppose D is a domain, and

1.10.23 F(z)=n=1an(z),
zD,

where an(z) is analytic for all n1, and the convergence of the product is uniform in any compact subset of D. Then F(z) is analytic in D.

If, also, an(z)0 when n1 and zD, then F(z)0 on D and

1.10.24 F(z)F(z)=n=1an(z)an(z).

Mittag-Leffler’s Expansion

If {an} and {zn} are sequences such that zmzn (mn) and n=1|anzn-2| is convergent, then

1.10.25 f(z)=n=1an(1z-zn+1zn)

is analytic in , except for simple poles at z=zn of residue an.