Chapter 16. Laurent Series. Residue Integration
2018학년도 2학기
반도체 공학과 공학수학2 (GEDB005) 강의교안 (2018학년도용)
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 최신공학수학 I 과 II, 한빛출판사 및 (영문) 다변량 Calculus (이상구, 김응기 et al) (http://www.hanbit.co.kr/EM/sage/)
강의시간: E12 (금 12:00 - 14:45), 반도체관 400126호 담당교수: 김응기 박사
주차 |
주교재 |
부교재 |
14 |
(if time permits) 16.1: Laurent Series 16.2: Singularities and Zeros. Infinity 16.3: Residue Integration Method 16.4: Residue Integration of Real Integrals |
15.1 복소 수열과 복소 무한급수 15.2 테일러 급수와 로랑급수 |
web |
Week 14
(If time permits)
Chapter 16 Laurent Series. Residue Integration
16.1. Laurent Series
Laurent Series generalize Taylor series.
is singular at we cannot use a Taylor series.
Using Laurent series.
We want to develop a function in powers of .
Laurent series consisting of positive integer powers of (and a constant) as well as negative integer powers of .
A Laurent series of converges in an annulus ( have singularities in the “hole”).
Theorem 1 Laurent’s Theorem
Let be analytic in a domain containing two concentric circles and with center and the annulus between them (blue in Fig. 370). Then can be represented by the Laurent series
(1)
consisting of nonnegative and negative powers. The coefficients of this Laurent series are given by the integrals
(2) ,
taken counterclockwise around any simple closed path that lies in the annulus and encircles the inner circle,[The variable of integration is denoted by since is used in ].
The series converges and represents in the enlarged open annulus obtained from the given annulus by continuously increasing the outer circle and decreasing until each of the two circles reaches a point where is singular.
In the important special case that is the only singular point of inside , this circle can be shrunk to the point , given convergence in a disk except at the center. In this case the series (or finite sum) of the negative powers of is called the principal part of the singularity of at .
Fig. 370. Laurent’s Theorem
Remark Instead of and we may write (denoting by )
(1')
where all the coefficients are now given by a single integral formula, namely
(2') .
Proof
(a) The non-negative powers are those of a Taylor series.
We use Cauchy's integral formula in Sec. 14.3 with *(instead of ) as the variable of integration and instead of .
Let and denote the functions represented by the two terms in in Sec. 14.3. Then
(3) .
Here is any point in the given annulus and we integrate counterclockwise over both and , so that the minus sign appears since in of Sec. the integration over is taken clockwise.
We transform each of these two integral as in Sec. .
The first integral is precisely as in Sec. .
Hence we get precisely the same result, namely, the Taylor series of ,
(4)
with coefficients [see , Sec , counterclockwise integration]
(5) .
Here we can replace by (see Fig. ), by the principle of deformation of path, since , the point where integrand in is not analytic, is not a point of the annulus.
This proves the formula for the in .
(b) The negative powers in and the formula for in are obtained if we consider (the second integral times in ).
Since lies in the annulus, it lies in the exterior of the path .
Hence the situation differs from that for the first integral.
The essential point is that instead of [see in Sec. ]
(6) we now have .
Consequently, we must develop the expression in the integrand of the second integral in in powers of (instead of the reciprocal of this) to get a convergent series.
We find
.
Compare this for a moment with in Sec. , to really understand the difference. Then go on and apply formula , Sec. , for a finite geometric sum, obtaining
.
Multiplication by and integration over both sides now yield
with the last term on the right given by
(7) .
As before, we can integrate over instead of in the integrals on the right. We see that on the right, the power is multiplied by as given in . This establishes Laurent’s theorem, provided
(8) .
(c) Convergence proof of . Very often will have only finitely many negative powers. Then there is nothing to be proved.
Otherwise, we begin by noting that in is bounded in absolute value, say,
for all on
because is analytic in the annuls and on , and lies on and outside, so that .
From this and the inequality applied to we get the inequality(length of , radius of const)
.
From we see that the expression on the right approaches zero as approaches infinity. This proves . The representation we coefficients is now established in the given annulus.
(d) Convergence of in the enlarged annulus.
The first series in is a Taylor series [representing ]; hence it converges in the disk with center whose radius equals the distance of the singularity(or singularities) closest to . Also, must be singular at all points outsides where is singular.
The second series in , representing , is a power series in .
Let the given annulus be where and are the radii of and respectively (Fig. 370). This corresponds to . Hence this power series in must converge at least in the disk . This corresponds to the exterior of , so that is analytic for all outside . Also, must be singular inside where is singular, and the series of the negative powers of (1) converges for all in the exterior E of the circle with center and radius equal to the maximum distance from to the singularities of inside . The domain common to and is the enlarged open annulus characterized near the end of Laurent’s theorem, whose proof is now complete.
Uniqueness.
The Laurent series of a given analytic function in its annulus of convergence is unique.
may have different Laurent series in two annuli with the same center.
The uniqueness is essential.
If a Laurent series has been found by any such process, the uniqueness guarantees that it must be the Laurent series of the given function in the given annulus.
Example 1 Use of Maclaurin Series
Find the Laurent series of with center .
Solution
By Maclaurin Series, we obtain
.
Here the “annulus” of convergence is the whole complex plane without the origin.
The principal part of the series at is .
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1/362880*z^4 - 1/5040*z^2 - 1/6/z^2 + 1/z^4 + 1/120
Example 2 Substitution
Find the Laurent series of with center .
Solution
By Maclaurin Series, we obtain
.
Example 3 Development of
Develop (a) in nonnegative powers of , (b) in negative powers of .
Solution
(a) (valid when )
(b)
(valid when )
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z^4 + z^3 + z^2 + z + 1
Example 4 Laurent Expansions in Different Concentric Annuli
Find all Laurent series of with center .
Solution
Multiplying by we get from Example 3
(I)
(II)
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z^4 + z^3 + z^2 + z + 1/z + 1/z^2 + 1/z^3 + 1
Example 5 Use of Partial Fractions
Find all Taylor and Laurent series of with center .
Solution
In terms of partial fractions,
.
For the first fraction.
(a)
(b)
For the second fraction,
(c)
(d)
(I) From (a) and (c), valid for (see Fig. 371),
.
Fig. 371. Regions of convergence in Example 5
(II) From (c) and (b), valid for
.
(III) From (d) and (b), valid for
.
If in Laurent’ theorem is analytic inside the coefficients in and are zero by Cauchy’s integral theorem, so that the Laurent series reduces to a Taylor series.
16.2 Singularities and Zeros. Infinity
A singular point of an analytic function is a at which ceases to be analytic.
A zero is a at .
We show that Laurent series can be used for classifying singularities and Taylor series for discussing zeros.
A function is singular or has a singularity at a point
if is not analytic (perhaps not even defined) at ,
but every neighborhood of contains points at which is analytic.
is a singular point of .
has a neighborhood without further singularities of
is an isolated singularity of
For example
has isolated singularities at , , etc.;
has a nonisolated singularity at .
Isolated singularities of at can be classified by the Laurent series
(1)
valid in the immediate neighborhood of the singular point , except at itself, that is, in a region of the form
.
The sum of the first series is analytic at .
The second series, containing the negative powers, is the principal part of Laurent series.
If it has only finitely many terms, it is of the form
(2) .
The singularity of at is a pole.
is called pole order.
Poles of the first order are simple poles.
If the principal part of Laurent series has infinitely many terms, has at an isolated essential singularity.
Example 1 Poles. Essential Singularities
The function
has a simple pole at and a pole of fifth order at .
Examples of functions having an isolated essential singularity at are
and
and have an isolated essential singularity at .
.
has a fourth-order pole at .
has a third-order pole at and a Laurent series with infinitely many negative powers.
This is no contradiction, since this series is valid for : it merely tells us that in classifying singularities it is quite important to consider the Laurent series valid in the immediate neighborhood of a singular point.
In Example 4 this is the series (I), which has three negative powers.
Example 2 Behavior Near a Pole
has a pole at and as in any manner.
Theorem 1 Poles
If is analytic and has a pole at , then as in any manner.
Example 3 Behavior Near an Essential Singularity
The function has an essential singularity at .
approach along the imaginary axis. does not exist.
if through positive real values.
if through negative real values.
Any given value in an arbitrarily small -neighborhood of .
We set
Nonzero complex number
Equating the absolute values and the arguments, we have ,
and
From these two equations and we obtain the formulas
and .
Hence can be made arbitrarily small by adding multiples of to leaving unaltered. This illustrates the very famous Picard’s theorem (with as the exceptional value).
Theorem 2 Picard’s Theorem
If is analytic and has an isolated essential singularity at a point , it takes on every value, with at most one exceptional value, in an arbitrarily small -neighborhood of .
Removable Singularities
A function has a removable singularity at if is not analytic at , but can be made analytic there by assigning a suitable value .
Example:
We define . is analytic at
Zeros of Analytic Functions
A zero of an analytic function in a domain is a in such that .
A zero has order , , , , are all at but .
A first-order zero is a simple zero.
For a second-order zero, but .
And so on.
Example 4 Zeros
The function has simple zeros at .
The function has second-order zeros at and .
The function has a third-order zero at .
The function has no zeros (see Sec. 13.5).
The function has simple zeros at , , , and has second-order zeros at these points.
The function has second-order zeros at , , , and the function has fourth-order zeros at these points.
Taylor Series at a Zero
At an -order zero of , the derivatives , , are zero.
Hence the first few coefficients , , of the Taylor series (1) are zero, whereas . Thus, this series takes the form
(3)
.
This is characteristic of such a zero, because, if has such a Taylor series, it has an -order zero at as follows by differentiation.
Theorem 3 Zeros
The zeros of an analytic function are isolated; that is, each of them has
a neighborhood that contains no further zeros of .
Proof
The factor in (3) is zero only at .
The power series in the brackets represents an analytic function (by Theorem 5 in Sec. 15.3), call it .
Now , since an analytic function is continuous, and because of this continuity, also in some neighborhood of .
Hence the same holds of .
Poles are often caused by zeros in the denominator. (Example: tan z has poles where
cos z is zero.) This is a major reason for the importance of zeros. The key to the connection is the following theorem, whose proof follows from (3) (see Team Project 12).
Theorem 4 Poles and Zeros
Let be analytic at and have a zero of order at . Then has a pole of order at ; and so does , provided is analytic at and .
Riemann Sphere. Point at Infinity
Riemann sphere.
A sphere of diameter touching the complex -plane at
Image of a point (a number z in the plane)
The intersection of the segment with , where is the “North Pole” diametrically opposite to the origin in the plane.
Each point on represents a complex number , except for , which does not correspond to any point in the complex plane.
Point at infinity (denoted “infinity”)
The image be .
Extended complex plane
The complex plane together with .
The complex plane is often called the finite complex plane, for distinction, or simply the complex plane as before.
The sphere S is called the Riemann sphere.
The mapping of the extended complex plane onto the sphere is known as a stereographic projection.
Analytic or Singular at Infinity
If we want to investigate a function for large ,
we may set and investigate in a neighborhood of . is analytic at infinity is analytic at .
is singular at infinity is singular at .
We also define
(4)
if this limit exists.
has an nth-order zero at infinity has such a zero at . Similarly for poles and essential singularities.
Example 5 Functions Analytic or Singular at Infinity.
Entire and Meromorphic Functions
The function is analytic at since is analytic at and has a second order zero at . The function is singular at and has a third-order pole there since the function has such a pole at . The function has an essential singularity at since has such a singularity at . Similarly, and have an essential singularity at .
Recall that an entire function is one that is analytic everywhere in the (finite) complex plane. Liouville’s
theorem (Sec. 14.4) tells us that the only bounded entire functions are the constants, hence any nonconstant entire function must be unbounded. Hence it has a singularity at , a pole if it is a polynomial or an essential singularity if it is not. The functions just considered are typical in this respect.
An analytic function whose only singularities in the finite plane are poles is called a meromorphic function.
Examples are rational functions with nonconstant denominator, , , and .
16.3. Residue Integration Method
The purpose of Cauchy’s residue integration method is the evaluation of integrals
taken around a simple closed path .
If is analytic everywhere on simple closed and inside , by Cauchy’s integral theorem .
If has a singularity at a point inside but is otherwise analytic on and inside . Then has a Laurent series
that converges for all points near (except at itself), in some domain of the form (sometimes called a deleted neighborhood, an old-fashioned term that we shall not use).
The coefficient of the first negative power of this Laurent series is given by the integral formula (2) in Sec. 16.1 with namely,
(1) . .
Here we integrate counterclockwise around a simple closed path that contains in its interior (but no other singular points of on or inside ).
The coefficient is the residue of at .
Example 1 Evaluation of an Integral by Means of a Residue
Integrate the function counterclockwise around the unit circle .
Solution
From (14) in Sec. 15.4 we obtain the Laurent series
which converges for (that is, for all ).
has a pole of third order at and the residue .
From (1) we thus obtain the answer
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-1/3*I*pi
Example 2 Use the Right Laurent Series!
Integrate clockwise around the circle : .
Solution
is singular at and .
Now lies outside . no interest
We need the residue of at .
We find it from the Laurent series that converges for .
This is series (I) in Example 4, Sec. 16.1,
.
This residue is 1. Clockwise integration thus yields
.
CAUTION! Had we used the wrong series (II) in Example 4, Sec. 16.1,
()
we would have obtained the wrong answer, , because this series has no power .
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-2*I*pi
Formulas for Residues
To calculate a residue at a pole, we need not produce a whole Laurent series, but, more economically, we can derive formulas for residues once and for all.
Simple Poles at .
A first formula for the residue at a simple pole is
(3) .
A second formula for the residue at a simple pole is
(4) .
In (4), with and has a simple zero at ,
has a simple pole at by Theorem 4 in Sec. 16.2.
Proof
We prove (3). For a simple pole at the Laurent series (1), Sec. 16.1, is
().
Here (Why?) Multiplying both sides by and then letting , we obtain
the formula (3):
where the last equality follows from continuity (Theorem 1, Sec. 15.3).
We prove (4). The Taylor series of at a simple zero is
.
Substituting this into and then into (3) gives
cancels. By continuity, the limit of the denominator is and (4) follows.
Example 3 Residue at a Simple Pole
has a simple pole at , and (3) gives the residue
.
By (4) with and we confirm the result,
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-5*I
Poles of Any Order at .
The residue of at an th-order pole at is
(5)
In particular, for a second-order pole ()
(5*)
Proof
We prove (5).
The Laurent series of converging near (except at itself) is (Sec. 16.2)
where . The residue wanted is . Multiplying both sides by gives
.
We see that is now the coefficient of the power of the power series of
. Hence Taylor’'s theorem (Sec. 15.4) gives (5):
.
Example 4 Residue at a Pole of Higher Order
has a pole of second order at .
From (5*) we obtain the residue
.
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8
Several Singularities Inside the Contour. Residue Theorem
Theorem 1 Residue Theorem
Let be analytic inside a simple closed path and on , except for finitely many singular points , , , inside . Then the integral of taken counterclockwise around equals times the sum of the residues of at , , , :
(6) .
Fig 373. Residue theorem
Example 5 Integration by the Residue Theorem. Several Contours
Evaluate the following integral counterclockwise around any simple closed path such that (a) and are inside , (b) is inside, outside, (c) is inside, outside, (d) and are outside.
Solution
The integrand has simple poles at and , with residues [by (3)]
.
.
[Confirm this by (4).] Answer:
(a)
(b)
(c)
(d)
Example 6 Another Application of the Residue Theorem
Integrate counterclockwise around the circle : .
Solution
is not analytic at , , , but all these points lie outside the contour . has simple poles at .
.
.
We thus obtain from (4) and the residue theorem
.
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1/2*tan(1)
1/2*tan(1)
Example 7 Poles and Essential Singularities
Evaluate the following integral, where is the ellipse (counterclockwise, sketch it).
.
Solution
Since at and ,
The first term of the integrand has simple poles at in inside , with residues [by (4); note that ]
.
.
.
has simple poles at which lie outside , so that they are of no interest here.
The second term of the integrand has an essential singularity at , with residue as obtained from
()
Answer: by the residue theorem.
16.4 Residue Integration of Real Integrals
Integrals of Rational Functions of and
(1)
where is a real rational function of and and is finite on the interval of integration.
Setting , we obtain
,
and
, , : counterclockwise
Example 1 An Integral of the Type (1)
Show by the present method that .
Solution
Using and .
The integral is
The integrand has a simple pole at outside the unit circle , so that it is of no interest here.
The integrand has a simple pole at inside the unit circle .
.
Answer:
Here is the factor in front of the last integral.
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-1/2
improper integral
(5) the Cauchy principal value of the integral.
Denote
.
It may exist even if the limits in do not.
Example:
We consider the corresponding contour integral
around a path C in Fig. 374.
Fig. 374. Path C of the contour integral in (5*)
is rational, has finitely many poles in the upper half-plane, and if we choose large enough, then encloses all these poles. By the residue theorem we then obtain
where the sum consists of all the residues of at the points in the upper half-plane at which has a pole. From this we have
(6) .
We prove that, if , the value of the integral over the semicircle approaches zero. If we set then is represented by , and as ranges along , the variable ranges from to . Since, by assumption, the degree of the denominator of is at least two units higher than the degree of the numerator, we have
()
for sufficiently large constants and . By the -inequality in Sec. 14.1,
()
Hence, as approaches infinity, the value of the integral over approaches zero, and (5) and (6) yield the result
(7)
where we sum over all the residues of at the poles of in the upper half-plane.
Example 2 An Improper Integral from 0 to
Using (7), show that
.
Solution
has four simple poles at the points (make a sketch)
, , ,
The first two of these poles lie in the upper half-plane (Fig. 375).
From (4) in the last section we find the residues
(Here we used and .) By (1) in Sec. 13.6 and (7) in this section,
.
Since is an even function, we thus obtain, as asserted,
.
Fourier Integrals
(10)
Example 3 An Application of (10)
Show that , (, )
Solution
has only one pole in the upper half-plane, namely, a simple pole at , and from (4) in Sec. 16.3 we obtain
.
Thus
.
Since , this yields the above results [see also (15) in Sec. 11.7.]
Another Kind of Improper Integral
This is called the Cauchy principal value of the integral. It is written
.
For example,
the principal value exists, although the integral itself has no meaning.
Theorem 1 Simple Poles on the Real Axis
If has a simple pole at on the real axis, then (Fig. 376)
.
where the first sum extends over all poles in the upper half-plane
the second over all poles on the real axis,
the latter being simple by assumption.
Example 4 Poles on the Real Axis
Find the principal value
Solution
Since
,
the integrand , considered for complex , has simple poles at
,
,
,
and at in the lower half-plane, which is of no interest here. From (14) we get the answer
.
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Contents
A. 공학수학 1 – 선형대수, 상미분방정식+ Lab
Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html
B. 공학수학 2 - 벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html
Copyright @ 2018 SKKU Matrix Lab. All rights reserved.
Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee
*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).