Chapter 16.   Laurent Series. Residue Integration


2018학년도 2학기

       반도체 공학과 공학수학2 (GEDB005) 강의교안  (2018학년도용)

    주교재: Erwin Kreyszig, Engineering Mathematics10th 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

http://www.hanbit.co.kr/EM/sage/2_chap15.html 

 


   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 .

                           그림입니다.

                      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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사각형입니다.

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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사각형입니다.

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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사각형입니다.

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),

.

                           그림입니다.

                 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

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 .

그림입니다.

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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사각형입니다.

-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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사각형입니다.

-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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사각형입니다.

-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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사각형입니다.

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)                     .

                       그림입니다.

                         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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사각형입니다.

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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사각형입니다.

-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.

            그림입니다.

    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

.

      그림입니다.

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)

   .

   그림입니다.

   

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).