Lecture #30: Laurent Series and Residue Theory

Transcription

1 Mathematics 312 (Fall 2013) November 21, 2012 Prof. Michael Kodron Lecture #30: Laurent Series and Residue Theory A first look at residue theory as an application of Laurent series One important application of the theory of Laurent series is in the computation of contour integrals. Suppose that f() isanalyticintheannulus0< 0 <Rso that f() hasa Laurent series expansion a j ( 0 ) j + a j ( 0 ) j. Let be any closed contour oriented counterclockwise lying entirely in the annulus and surrounding 0 so that f()d = a j We know from Theorem 23.2 that so that ( 0 ) j d = ( 0 ) j d + a j 2πi, if j = 1, 0, if j = 1, f()d =2πia 1. ( 0 ) j d. Thus, we see that the coefficient a 1 in the Laurent series expansion of f() inanannulus of the form 0 < 0 <Ris of particular importance. In fact, it has a name! Definition. Suppose that the function f() hasanisolatedsingularityat 0.Thecoefficient a 1 of ( 0 ) 1 in the Laurent series expansion of f() around 0 is called the residue of f() at 0 and is denoted by a 1 =Res(f; 0 ). lassifying isolated singularities We will now focus on functions that have an isolated singularity at 0. Therefore, suppose that f() isanalyticintheannulus0< 0 <Rand has an isolated singularity at 0. onsider its Laurent series expansion a j ( 0 ) j + a j ( 0 ) j. 30 1

2 We call the part with the negative powers of ( 0 ), namely a j ( 0 ) j, the principal part of the Laurent series. There are three mutually exclusive possibilities for the principal part. (i) If a j =0forallj<0, then we say that 0 is a removable singularity of f(). (ii) If a m =0forsomem N, buta j =0forallj< m, thenwesaythat 0 is a pole of order m for f(). (iii) If a j =0forinfinitelymanyj<0, then we say that 0 is an essential singularity of f(). Example Suppose that sin for > 0. Since the Laurent series expansion of f() is 1 sin = 1 3 3! + 5 5! 7 7! + we conclude that 0 =0isaremovablesingularity. Example Suppose that e m for > 0wherem is a positive integer. Since 1 m ! + 3 3! + =1 2 3! + 4 5! 6 7! +, = 1 m + 1 m ! m (m 1)! + 1 m! + (m +1)! +, we conclude that 0 =0isapoleoforderm. Example Suppose that for > 0. Since e 1/ e 1/ =1+(1/)+ (1/)2 2! + (1/)3 3! we conclude that 0 =0isanessentialsingularity. + = ! ! 3 +, 30 2

3 Example Suppose that = { =1} denotes the unit circle oriented counterclockwise. ompute the following three integrals: (a) (b) (c) sin d, e d, wherem is a positive integer, and m e 1/ d. Solution. In order to compute all three integrals, we use the fact that 0 =0isanisolated singularity so that f()d =2πi Res(f;0). (a) From Example 30.1, we know that a 1 =0sothat sin d =0. (b) From Example 30.2, we know that a 1 =1/(m 1)! so that e d = 2πi m (m 1)!. (c) From Example 30.3, we know that a 1 =1sothat e 1/ d =2πi. Note that although we could have used the auchy Integral Formula to solve (a) and (b), we could not have used it to solve (c). Question. Given the obvious importance of the coefficient a 1 in a Laurent series, it is natural to ask if there is any way to determine a 1 without computing the entire Laurent series. Theorem A function f() has a pole of order m at 0 if and only if g() ( 0 ) m for some function g() that is analytic in a neighbourhood of 0 and has g( 0 ) =

4 Proof. Suppose that f() hasapoleoforderm at 0.Bydefinition,theLaurentseriesfor f() hastheform a m ( 0 ) + a m j ( 0 ) j and so ( 0 ) m a m + Therefore, if we let j= (m 1) g() =a m + j= (m 1) a j ( 0 ) j+m = a m + a j m ( 0 ) j, a j m ( 0 ) j. then g() isanalyticinaneighbourhoodof 0. By assumption, a m =0sincef() hasa pole of order m at 0,andsog( 0 )=a m =0. onversely, suppose that g() ( 0 ) m for some function g() thatisanalyticinaneighbourhoodof 0 and has g( 0 ) = 0. Since g() isanalytic,itcanbeexpandedinataylorseriesabout 0,say g() =b 0 + b 1 ( 0 )+b 2 ( 0 ) 2 + = b j ( 0 ) j. Since g( 0 )=b 0 =0byassumption,weobtain 1 ( 0 ) m b j ( 0 ) j = Therefore, by definition, f() hasapoleoforderm at 0. b 0 ( 0 ) + b 1 +. m ( 0 ) m 1 Now that we know the general form of a function f() thathasapoleoforderm at 0,we can determine a formula for Res(f; 0 ), the residue of f() at 0,asfollows. Suppose that f() hasapoleoforderm at 0 so that from Theorem 30.5 we have g() ( 0 ) m for some function g() thatisanalyticinaneighbourhoodof 0 and has g( 0 ) = 0. If is aclosedcontourorientedcounterclockwisecontaining 0 and f() isanalyticinsideandon except for a pole of order m at 0,thenfromourLaurentseriesdevelopmentwehave f()d =2πi Res(f; 0 ). ( ) 30 4

5 On the other hand, we can apply the auchy Integral Formula to conclude so equating ( ) and( ) implies f()d = g() =2πig(m 1) ( 0 ) d ( 0 ) m (m 1)!, ( ) Res(f; 0 )= g(m 1) ( 0 ) (m 1)!. Since g() =( 0 ) m f(), we conclude 1 d m 1 Res(f; 0 )= (m 1)! d ( 0) m f() m 1. =0 Theorem If f() is analytic for 0 < 0 <Rand has a pole of order m at 0, then 1 d m 1 Res(f; 0 )= (m 1)! d ( 0) m f() 1 m 1 = =0 (m 1)! lim d m 1 0 d ( 0) m f(). m 1 In particular, if 0 is a simple pole, then Res(f; 0 )=( 0 )f() =lim( 0 )f(). 0 =0 30 5