Baixe Arfken cap7 e outras Manuais, Projetos, Pesquisas em PDF para Física, somente na Docsity!
Chapter 7
Functions of a
Complex Variable II
Calculus of Residues
7.1 Singularities
In this chapter we return to the line of analysis that started with the Cauchy– Riemann conditions in Chapter 6 and led to the Laurent expansion (Sec- tion 6.5). The Laurent expansion represents a generalization of the Taylor series in the presence of singularities. We define the point z 0 as an isolated singular point of the function f ( z ) if f ( z ) is not analytic at z = z 0 but is analytic and single valued in a punctured disk 0 < | z − z 0 | < R for some positive R. For rational functions, which are quotients of polynomials, f ( z ) = P ( z )/ Q ( z ), the only singularities arise from zeros of the denominator if the numerator is nonzero there. For example, f ( z ) = z
(^3) − 2 z (^2) + 1 ( z −3)( z^2 +3) from Exercise 6.5.13 has simple poles at z = ± i
3 and z = 3 and is regular everywhere else. A function that is analytic throughout the finite complex plane except for isolated poles is called meromorphic. Examples are entire functions that have no singularities in the finite complex plane, such as e z , sin z , cos z , rational functions with a finite number of poles, or tan z , cot z with infinitely many isolated simple poles at z = n π and z = (2 n +1)π/2 for n = 0, ±1, ±2,... , respectively. From Cauchy’s integral we learned that a loop integral of a function around a simple pole gives a nonzero result, whereas higher order poles do not con- tribute to the integral (Example 6.3.1). We consider in this chapter the general- ization of this case to meromorphic functions leading to the residue theorem, which has important applications to many integrals that physicists and engi- neers encounter, some of which we will discuss. Here, singularities, and simple poles in particular, play a dominant role.
372
7.1 Singularities 373
Poles
In the Laurent expansion of f ( z ) about z 0
f ( z ) =
n =−∞
a (^) n ( z − z 0 ) n , (7.1)
if a (^) n = 0 for n < − m < 0 and a − m = 0, we say that z 0 is a pole of order m. For instance, if m = 1—that is, if a − 1 /( z − z 0 ) is the first nonvanishing term in the Laurent series—we have a pole of order 1, often called a simple pole. Example 6.5.1 is a relevant case: The function
f ( z ) = [ z ( z − 1)]−^1 = −
z
z − 1 has a simple pole at the origin and at z = 1. Its square, f^2 ( z ), has poles of order 2 at the same places and [ f ( z )] m^ has poles of order m = 1, 2, 3.... In contrast, the function
e
(^1) z
n = 0
z n^ n! from Example 6.5.2 has poles of any order at z = 0. If there are poles of any order (i.e., the summation in the Laurent series at z 0 continues to n = −∞), then z 0 is a pole of infinite order and is called an essential singularity. These essential singularities have many pathological features. For instance, we can show that in any small neighborhood of an essential singularity of f ( z ) the function f ( z ) comes arbitrarily close to any (and therefore every) preselected complex quantity w 0. 1 Literally, the entire w- plane is mapped into the neighborhood of the point z 0 , the essential singularity. One point of fundamental difference between a pole of finite order and an essential singularity is that a pole of order m can be removed by multiplying f ( z ) by ( z − z 0 ) m. This obviously cannot be done for an essential singularity. The behavior of f ( z ) as z → ∞ is defined in terms of the behavior of f (1/ t ) as t → 0. Consider the function
sin z =
n = 0
(−1) n^ z^2 n +^1 (2 n + 1)!
As z → ∞, we replace the z by 1/ t to obtain
sin
t
n = 0
(−1) n (2 n + 1)! t^2 n +^1
Clearly, from the definition, sin z has an essential singularity at infinity. This result could be anticipated from Exercise 6.1.9 since
sin z = sin iy = i sinh y , when x = 0,
(^1) This theorem is due to Picard. A proof is given by E. C. Titchmarsh, The Theory of Functions , 2nd ed. Oxford Univ. Press, New York (1939).
7.1 Singularities 375
The first factor on the right-hand side, ( z + 1)^1 /^2 , has a branch point at z = −1. The second factor has a branch point at z = +1. Each branch point has order 2 because the Riemann surface is made up of two sheets. At infinity f ( z ) has a simple pole. This is best seen by substituting z = 1 / t and making a binomial expansion at t = 0
( z^2 − 1)^1 /^2 =
t
(1 − t^2 )^1 /^2 =
t
∑^ ∞
n = 0
n
(−1) n^ t^2 n^ =
t
t −
t^3 + · · ·.
The cut line has to connect both branch points so that it is not possi- ble to encircle either branch point completely. To check the possibility of taking the line segment joining z = +1 and z = −1 as a cut line, let us fol- low the phases of these two factors as we move along the contour shown in Fig. 7.1.
For convenience in following the changes of phase, let z + 1 = re i θ^ and z − 1 = ρ e i ϕ^. Then the phase of f ( z ) is (θ + ϕ)/2. We start at point 1, where both z + 1 and z − 1 have a phase of zero. Moving from point 1 to point 2, ϕ, the phase of z − 1 = ρ e i ϕ^ increases by π. ( z − 1 becomes negative.) ϕ then stays constant until the circle is completed, moving from 6 to 7. θ, the phase of z + 1 = re i θ^ , shows a similar behavior increasing by 2π as we move from 3 to 5. The phase of the function f ( z ) = ( z + 1)^1 /^2 ( z − 1)^1 /^2 = r^1 /^2 ρ^1 /^2 e i (θ+ϕ)/^2 is (θ + ϕ)/2. This is tabulated in the final column of Table 7.1.
Table 7.
Phase Angle
Point θ ϕ ( θ + ϕ )/
1 0 0 0 2 0 π π/ 2 3 0 π π/ 2 4 π π π 5 2 π π 3 π/ 2 6 2 π π 3 π/ 2 7 2 π 2 π 2 π
y
x
3
5
(^4) − 1 2 +
6
1
7
z
Figure 7. Cut Line Joining Two Branch Points at ± 1
376 Chapter 7 Functions of a Complex Variable II
− 1 +
j > 0
j < 0 (a)
− (^1) + 1
−∞ +∞
(b)
Figure 7.
Branch Points Joined by (a) a Finite Cut Line and (b) Two Cut Lines from 1 to ∞ and − 1 to −∞ that Form a Single Cut Line Through the Point at Infinity. Phase Angles are Measured as Shown in (a) Two features emerge:
- The phase at points 5 and 6 is not the same as the phase at points 2 and 3. This behavior can be expected at a branch point cut line.
- The phase at point 7 exceeds that at point 1 by 2π and the function f ( z ) = ( z^2 − 1)^1 /^2 is therefore single-valued for the contour shown, encircling both branch points.
If we take the x -axis − 1 ≤ x ≤ 1 as a cut line, f ( z ) is uniquely specified. Alternatively, the positive x -axis for x > 1 and the negative x -axis for x < − 1 may be taken as cut lines. In this case, the branch points at ±1 are joined by the cut line via the point at infinity. Again, the branch points cannot be encircled and the function remains single-valued. ■
Generalizing from this example, the phase of a function
f ( z ) = f 1 ( z ) · f 2 ( z ) · f 3 ( z ) · · ·
is the algebraic sum of the phase of its individual factors:
arg f ( z ) = arg f 1 ( z ) + arg f 2 ( z ) + arg f 3 ( z ) + · · ·.
The phase of an individual factor may be taken as the arctangent of the ratio of its imaginary part to its real part,
arg fi ( z ) = tan−^1 (v i / ui ),
but one should be aware of the different branches of arctangent. For the case of a factor of the form
fi ( z ) = ( z − z 0 ),
the phase corresponds to the phase angle of a two-dimensional vector from
- z 0 to z , with the phase increasing by 2π as the point + z 0 is encircled provided it is measured without crossing a cut line ( z 0 = 1 in Fig. 7.2a). Conversely, the traversal of any closed loop not encircling z 0 does not change the phase of z − z 0. Poles are the simplest singularities, and functions that have only poles besides regular points are called meromorphic. Examples are tan z and ratios
378 Chapter 7 Functions of a Complex Variable II
7.2 Calculus of Residues
Residue Theorem
If the Laurent expansion of a function f ( z ) =
n =−∞ a^ n ( z^ −^ z^0 ) n (^) is integrated term by term by using a closed contour that encircles one isolated singular point z 0 once in a counterclockwise sense, we obtain (Example 6.3.1)
a (^) n
( z − z 0 ) n^ dz = a (^) n
( z − z 0 ) n +^1 n + 1
z 1
z 1
= 0, n = − 1. (7.5)
However, if n = −1, using the polar form z = z 0 + re i θ^ we find that
a − 1
( z − z 0 )−^1 dz = a − 1
ire i θ^ d θ re i θ^
= 2 π ia − 1. (7.6)
The first and simplest case of a residue occurred in Example 6.3.1 involving∮ z n^ dz = 2 π ni δ n ,− 1 , where the integration is anticlockwise around a circle of radius r. Of all powers z n , only 1/ z contributes. Summarizing Eqs. (7.5) and (7.6), we have 1 2 π i
f ( z ) dz = a − 1. (7.7)
The constant a − 1 , the coefficient of ( z − z 0 )−^1 in the Laurent expansion, is called the residue of f ( z ) at z = z 0. A set of isolated singularities can be handled by deforming our contour as shown in Fig. 7.3. Cauchy’s integral theorem (Section 6.3) leads to ∮
C
f ( z ) dz +
C 0
f ( z ) dz +
C 1
f ( z ) dz +
C 2
f ( z ) dz + · · · = 0. (7.8)
C
C 0
C 1
C 2
ℑ z = y
ℜ z = x
Figure 7. Excluding Isolated Singularities
7.2 Calculus of Residues 379
The circular integral around any given singular point is given by Eq. (7.7), ∮
C (^) i
f ( z ) dz = − 2 π ia − 1 zi , (7.9)
assuming a Laurent expansion about the singular point z = zi. The negative sign comes from the clockwise integration as shown in Fig. 7.3. Combining Eqs. (7.8) and (7.9), we have ∮
C
f ( z ) dz = 2 π i ( a − 1 z 0 + a − 1 z 1 + a − 1 z 2 + · · ·)
= 2 π i (sum of enclosed residues). (7.10)
This is the residue theorem. The problem of evaluating one or more con- tour integrals is replaced by the algebraic problem of computing residues at the enclosed singular points (poles of order 1). In the remainder of this section, we apply the residue theorem to a wide variety of definite integrals of mathemati- cal and physical interest. The residue theorem will also be needed in Chapter 15 for a variety of integral transforms, particularly the inverse Laplace transform. We also use the residue theorem to develop the concept of the Cauchy principal value. Using the transformation z = 1 /w for w → 0, we can find the nature of a singularity at z → ∞ and the residue of a function f ( z ) with just isolated singularities and no branch points. In such cases, we know that ∑ {residues in the finite z -plane} + {residue at z → ∞} = 0.
Evaluation of Definite Integrals
The calculus of residues is useful in evaluating a wide variety of definite inte- grals in both physical and purely mathematical problems. We consider, first, integrals of the form
I =
∫ (^2) π
0
f (sin θ, cos θ) d θ, (7.11)
where f is finite for all values of θ. We also require f to be a rational function of sin θ and cos θ so that it will be single-valued. Let
z = e i θ^ , dz = ie i θ^ d θ.
From this,
d θ = − i
dz z
, sin θ =
z − z −^1 2 i
, cos θ =
z + z −^1 2
Our integral becomes
I = − i
f
z − z −^1 2 i
z + z −^1 2
dz z
7.2 Calculus of Residues 381
y
z
x − R R
R → ∞
Figure 7.
With these conditions, we may take as a contour of integration the real axis and a semicircle in the upper half-plane as shown in Fig. 7.4. We let the radius R of the semicircle become infinitely large. Then ∮ f ( z ) dz = lim R →∞
∫ R
− R
f ( x ) dx + lim R →∞
∫ (^) π
0
f ( Re i θ^ ) iRe i θ^ d θ
= 2 π i
residues (upper half-plane) (7.16)
From the second condition, the second integral (over the semicircle) vanishes and ∫ (^) ∞
−∞
f ( x ) dx = 2 π i
residues (upper half-plane). (7.17)
Note that a corresponding result is obtained when f is analytic in the lower half-plane and we use a contour in the lower half-plane. In that case, the contour will be tracked clockwise and the residues will enter with a minus sign.
EXAMPLE 7.2.2 Inverse Polynomial^ Evaluate
I =
−∞
dx 1 + x^2
From Eq. (7.16), ∫ (^) ∞
−∞
dx 1 + x^2
= 2 π i
residues (upper half-plane).
Here and in every other similar problem, we have the question: Where are the poles? Rewriting the integrand as 1 z^2 + 1
z + i
z − i
we see that there are simple poles (order 1) at z = i and z = − i.
382 Chapter 7 Functions of a Complex Variable II
A simple pole at z = z 0 indicates (and is indicated by) a Laurent expansion of the form
f ( z ) =
a − 1 z − z 0
n = 1
a (^) n ( z − z 0 ) n. (7.20)
The residue a − 1 is easily isolated as (Exercise 7.1.1)
a − 1 = ( z − z 0 ) f ( z ) | z = z 0. (7.21)
Using Eq. (7.21), we find that the residue at z = i is 1/ 2 i , whereas that at z = − i is − 1 / 2 i. Another way to see this is to write the partial fraction decomposition: 1 z^2 + 1
2 i
z − i
z + i
Then ∫ (^) ∞
−∞
dx 1 + x^2
= 2 π i ·
2 i
= π. (7.22)
Here, we used a − 1 = 1 / 2 i for the residue of the one included pole at z = i. Readers should satisfy themselves that it is possible to use the lower semicircle and that this choice will lead to the same result: I = π. ■
A more delicate problem is provided by the next example.
EXAMPLE 7.2.3 Evaluation of Definite Integrals^ Consider^ definite integrals of the form
I =
−∞
f ( x ) e iax^ dx , (7.23)
with a real and positive. This is a Fourier transform (Chapter 15). We assume the two conditions:
- (^) f ( z ) is analytic in the upper half-plane except for a finite number of poles.
- (^) lim | z |→∞ f ( z ) = 0, 0 ≤ arg z ≤ π. (7.24)
Note that this is a less restrictive condition than the second condition imposed on f ( z ) for integrating
−∞ f^ ( x )^ dx.
We employ the contour shown in Fig. 7.4 because the exponential factor goes rapidly to zero in the upper half-plane. The application of the calculus of residues is the same as the one just considered, but here we have to work harder to show that the integral over the (infinite) semicircle goes to zero. This integral becomes
I R =
∫ (^) π
0
f ( Re i θ^ ) e iaR^ cos^ θ− aR^ sin^ θ^ iRe i θ^ d θ. (7.25)
384 Chapter 7 Functions of a Complex Variable II
x 0
Figure 7. Bypassing Singular Points
Since the integral over the upper semicircle I (^) R vanishes as R → ∞ (Jordan’s lemma), ∫ (^) ∞
−∞
f ( x ) e iax^ dx = 2 π i
residues(upper half-plane) ( a > 0). (7.30)
This result actually holds more generally for complex a with ( a ) > 0. ■
Cauchy Principal Value
Occasionally, an isolated first-order pole will be directly on the contour of integration. In this case, we may deform the contour to include or exclude the residue as desired by including a semicircular detour of infinitesimal radius. This is shown in Fig. 7.6. The integration over the semicircle then gives, with z − x 0 = δ e i ϕ^ , dz = i δ e i ϕ^ d ϕ, ∫ dz z − x 0
= i
∫ (^2) π
π
d ϕ = i π, i.e., π ia − 1 if counterclockwise, ∫ dz z − x 0
= i
π
d ϕ = − i π, i.e., − π ia − 1 if clockwise.
This contribution, + or −, appears on the left-hand side of Eq. (7.10). If our detour were clockwise, the residue would not be enclosed and there would be no corresponding term on the right-hand side of Eq. (7.10). However, if our detour were counterclockwise, this residue would be enclosed by the contour C and a term 2π ia − 1 would appear on the right-hand side of Eq. (7.10). The net result for either clockwise or counterclockwise detour is that a simple pole on the contour is counted as one-half what it would be if it were within the contour. For instance, let us suppose that f ( z ) with a simple pole at z = x 0 is inte- grated over the entire real axis assuming | f ( z )| → 0 for | z | → ∞ fast enough (faster than 1/| z |) that the integrals in question are finite. The contour is closed with an infinite semicircle in the upper half-plane (Fig. 7.7). Then ∮ f ( z ) dz =
∫ (^) x 0 −δ
−∞
f ( x ) dx +
C (^) x 0
f ( z ) dz
x 0 +δ
f ( x ) dx +
C
infinite semicircle
= 2 π i
enclosed residues. (7.31)
7.2 Calculus of Residues 385
x 0
y
z
x
Figure 7.
Closing the Contour with an Infinite Radius Semicircle
If the small semicircle C (^) x 0 includes x 0 (by going below the x -axis; counter- clockwise),∫ x 0 is enclosed, and its contribution appears twice —as π ia − 1 in
C (^) x 0 and as 2π ia −^1 in the term 2π i^
enclosed residues—for a net contribution of π ia − 1 on the right-hand side of Eq. (7.31). If the upper small semicircle is se- lected, x 0 is excluded. The only contribution is from the clockwise integration over C (^) x 0 , which yields −π ia − 1. Moving this to the extreme right of Eq. (7.11), we have +π ia − 1 , as before. The integrals along the x -axis may be combined and the semicircle radius permitted to approach zero. We therefore define
lim δ→ 0
{∫ (^) x 0 −δ
−∞
f ( x ) dx +
x 0 +δ
f ( x ) dx
= P
−∞
f ( x ) dx. (7.32)
P indicates the Cauchy principal value and represents the preceding limit- ing process. Note that the Cauchy principal value is a balancing or canceling process; for even-order poles, P
−∞ f^ ( x )^ dx^ is not finite because there is no cancellation. In the vicinity of our singularity at z = x 0 ,
f ( x ) ≈
a − 1 x − x 0
This is odd, relative to x 0. The symmetric or even interval (relative to x 0 ) provides cancellation of the shaded areas (Fig. 7.8). The contribution of the singularity is in the integration about the semicircle. Sometimes, this same limiting technique is applied to the integration limits ±∞. If there is no singularity, we may define
P
−∞
f ( x ) dx = lim a →∞
∫ (^) a
− a
f ( x ) dx. (7.34)
An alternate treatment moves the pole off the contour and then considers the limiting behavior as it is brought back, in which the singular points are moved off the contour in such a way that the integral is forced into the form desired to satisfy the boundary conditions of the physical problem (for Green’s functions this is often the case; see Examples 7.2.5 and 8.11.2). The principal value limit
7.2 Calculus of Residues 387
C 1
C 2
r (^) R x
Figure 7. Singularity on Contour
the final zero coming from the residue theorem [Eq. (7.10)]. By Jordan’s lemma, ∫
C 2
e iz^ dz z
and ∮ e iz^ dz z
C 1
e iz^ dz z
+ P
−∞
e ix^ dx x
The integral over the small semicircle yields (−)π i times the residue of 1, the minus as a result of going clockwise. Taking the imaginary part, 5 we have ∫ (^) ∞
−∞
sin x x
dx = π (7.40)
or ∫ (^) ∞
0
sin x x
dx =
π 2
The contour of Fig. 7.9, although convenient, is not at all unique. Another choice of contour for evaluating Eq. (7.35) is presented as Exercise 7.2.15. ■
EXAMPLE 7.2.5 Quantum Mechanical Scattering^ The quantum mechanical analysis of scattering leads to the function
I (σ ) =
−∞
x sin xdx x^2 − σ 2
where σ is real and positive. From the physical conditions of the problem there is a further requirement: I (σ ) must have the form e i σ^ so that it will represent an outgoing scattered wave.
(^5) Alternatively, we may combine the integrals of Eq. (7.37) as ∫ (^) − r − R
e ix^ dx x +
∫ (^) R r
e ix^ dx x =
∫ (^) R r
( e ix^ − e − ix ) dx x =^2 i
∫ (^) R r
sin x x dx.
388 Chapter 7 Functions of a Complex Variable II
y
I 1
x − s − s
I 1
(a)
y
I 2
x
(b)
s I 2 s
Figure 7.
Contours
Using
sin z =
2 i
e iz^ −
2 i
e − iz , (7.43)
we write Eq. (7.42) in the complex plane as
I (σ ) = I 1 + I 2 , (7.44)
with
I 1 =
2 i
−∞
ze iz z^2 − σ 2
dz ,
I 2 = −
2 i
−∞
ze − iz z^2 − σ 2
dz. (7.45)
Integral I 1 is similar to Example 7.2.2 and, as in that case, we may complete the contour by an infinite semicircle in the upper half-plane as shown in Fig. 7.10a. For I 2 , the exponential is negative and we complete the contour by an infinite semicircle in the lower half-plane, as shown in Fig. 7.10b. As in Example 7.2.2, neither semicircle contributes anything to the integral–Jordan’s lemma.
There is still the problem of locating the poles and evaluating the residues. We find poles at z = +σ and z = −σ on the contour of integration. The residues are (Exercises 7.1.1 and 7.2.1)
z = σ z = −σ
I 1
e i σ 2
e − i σ 2
I 2
e − i σ 2
e i σ 2
390 Chapter 7 Functions of a Complex Variable II
the Cauchy principal value of the integral. Note that we have these possi- bilities [Eqs. (7.48), (7.52), and (7.53)] because our integral is not uniquely defined until we specify the particular limiting process (or average) to be used. ■
Pole Expansion of Meromorphic Functions
Analytic functions f ( z ) that have only separated poles as singularities are called meromorphic. Examples are ratios of polynomials, cot z [see (^) dzd ln sin z of Eq. (5.139)] and f^
′( z ) f ( z ) of entire functions. For simplicity, we assume that these poles at finite z = zn with 0 < | z 1 | < | z 2 | < · · · are all simple, with residues bn. Then an expansion of f ( z ) in terms of b (^) n ( z − zn )−^1 depends only on intrinsic properties of f ( z ), in contrast to the Taylor expansion about an arbitrary analytic point of f ( z ) or the Laurent expansion about some singular point of f ( z ).
EXAMPLE 7.2.6 Rational Functions^ Rational functions are ratios of polynomials that can be completely factored according to the fundamental theorem of algebra. A partial fraction expansion then generates the pole expansion. Let us consider a few simple examples. We check that the meromorphic function
f ( z ) =
z ( z + 1)
z
z + 1
has simple poles at z = 0, z = −1 with residues (^) z ( zz +1) | z = 0 = (^) z +^11 | z = 0 = 1 and z + 1 z ( z +1) | z =−^1 =^
1 z = −^ 1, respectively. At^ ∞,^ f^ ( z ) has a second order zero. Simi- larly, the meromorphic function
f ( z ) =
z^2 − 1
z − 1
z + 1
has simple poles at z = ±1 with residues (^) zz 2 −−^11 | z = 1 = (^) z +^11 | z = 1 = 12 and (^) zz 2 +−^11 | z =− 1 = 1 z − 1 | z =−^1 = −^
1 2.^ At infinity,^ f^ ( z ) has a second-order zero also.^ ■
Let us consider a series of concentric circles C (^) n about the origin so that C (^) n includes z 1 , z 2 ,... , zn but no other poles, its radius R (^) n → ∞ as n → ∞. To guarantee convergence we assume that | f ( z )| < ε R (^) n for an arbitrarily small positive constant ε and all z on C (^) n , and f is regular at the origin. Then the series
f ( z ) = f (0) +
n = 1
bn
( z − zn )−^1 + z − n^1
converges to f ( z ). To prove this theorem (due to Mittag–Leffler) we use the residue theorem to evaluate the following contour integral for z inside C (^) n and not equal to a
7.2 Calculus of Residues 391
singular point of f (w)/w:
In =
2 π i
C (^) n
f (w) w(w − z )
d w
∑^ n
m = 1
bm z (^) m ( z (^) m − z )
f ( z ) − f (0) z
where w in the denominator is needed for convergence and w − z to produce f ( z ) via the residue theorem. On C (^) n we have for n → ∞,
| In | ≤ 2 π R (^) n
maxw on C (^) n | f (w)| 2 π R (^) n ( R (^) n − | z |)
ε R (^) n R (^) n − | z |
≤ ε
for R (^) n | z | so that | In | → 0 as n → ∞. Using In → 0 in Eq. (7.55) proves Eq. (7.54). If | f ( z )| < ε R p n +^1 for some posive integer p , then we evaluate similarly the integral
In =
2 π i
f (w) w p +^1 (w − z )
d w → 0 as n → ∞
and obtain the analogous pole expansion
f ( z ) = f (0) + zf ′(0) + · · · +
z p^ f ( p )^ (0) p!
n = 1
b (^) n z p +^1 / z (^) np +^1 z − z (^) n
Note that the convergence of the series in Eqs. (7.54) and (7.56) is implied by the bound of | f ( z )| for | z | → ∞.
EXAMPLE 7.2.7 Pole Expansion of Cotangent^ The meromorphic function^ f^ ( z )^ =^ π^ cot^ π^ z has simple poles at z = ± n , for n = 0, 1, 2,... with residues (^) d π sin^ cos π z π/^ zdz | z = n = π cos π n π cos π n =^1.^ Before we apply the Mittag–Leffler theorem, we have to subtract the pole at z = 0. Then the pole expansion becomes
π cot π z −
z
n = 1
z − n
n
z + n
n
n = 1
z − n
z + n
n = 1
2 z z^2 − n^2
We check this result by taking the logarithm of the product for the sine [Eq. (7.60)] and differentiating
π cot π z =
z
n = 1
n (1 − zn )
n (1 + zn )
z
n = 1
z − n
z + n
Finally, let us also compare the pole expansion of the rational functions of Example 7.2.6 with the earlier partial fraction forms. From the Mittag–Leffler theorem, we get 1 z^2 − 1
z − 1
z + 1
