EE2012 ~ Page 9 / Part 2. ben m chen, nus ece

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1 omplex Analysis EE ~ Page 9 / Part

2 Flow hart of Material in omplex Analysis x iy t () xt () iyt () f( ) uivu( x, y) iv( x, y) Starting omplex function of a real variable ~ define a curve on a complex plane omplex function of a complex variable ~ define a complex mapping from one complex plane to another f ( ) d f ( ) d f () t () t dt auchy s Integral theorem ~ valid when f () is analytical for every point encircled by a closed curve ~ auchy Riemann equations, singularities omplex integral ~ integration of a complex function (of a complex variable) on a complex curve ~ upper bound of a complex integration to be continued on the next page EE ~ Page / Part

3 Flow hart of Material in omplex Analysis (cont.) f ( ) f( ) d i f ( ) ( ) n d i f ( n) ( ) n! auchy s integral formula ~ is a closed curve encloses and f () is analytic for every point inside Another integral formula ~ n, is a closed curve encloses and f () is analytic inside ~ power series of an analytic function Applications of complex integral to be continued on the next page f ( ) d i Res ( f, ) complex integral in terms of residues ~ Taylor series, Laurent series, order of singularities, residues ~ The most general result for complex integral! EE ~ Page / Part

4 Flow hart of Material in omplex Analysis (cont.) Applications of complex integral f (cos,sin ) d f ( x ) dx Px ( ) ( ) Qx dx f xcos mxdx f xsin mxdx Laplace transform argument principle Rouche s theorem Mission completed EE ~ Page / Part

5 Basic Operations of omplex Numbers artesian and Polar oordinates: y i tan iarg x x iy e x y e cos(arg ) i sin(arg ) Euler s Formula: i e cos( ) isin( ) Additions: It is easy to do additions (subtractions) in artesian coordinate, i.e., ( a ib) ( viw) ( av) i( bw) Multiplication: It is easy to do multiplication (division) in Polar coordinate, i.e., i i i i( ) i( ) ( ) i re ue ru e re ue r e u EE ~ Page 3 / Part

6 omplex Functions of a Real Variable omplex functions of a real variable are needed to represent paths or contours in the complex plane. t () xt () iyt (), t[ ab, ] Example 5 it 3 t () 5 e, t [, ] 5costi5sint 4 xt () 5cost yt () 5sin, t t[, ] EE ~ Page 4 / Part

7 Properties of omplex Function of Real Variable lim t ( ) lim xt ( ) i lim y( t) ta ta ta is continuous if x and y are continuous, i.e. lim xt ( ) xa ( ), lim yt ( ) ya ( ) () t x() t i y() t (t) is smooth if () t is continuous, i.e. if x () t and y() t are continuous. ta ta (t) is piecewise smooth if (t) is smooth everywhere except for a finite number of discontinuities. EE ~ Page 5 / Part

8 Properties of omplex Function of Real Variable (cont.) Normal differentiation and integration rules are applicable: ( c c) cc b b b ( c c ) dt c dt c dt a a a b dt b ( ) a ( ) a EE ~ Page 6 / Part

9 urves The set of images {() t t[, a b]} is called a curve in the complex plane 4 Smooth curve Smooth Smooth closed closed curve curve - - The curve is smooth if Piecewise smooth curve Piecewise smooth (pws) curve () t is continuous - EE ~ Page 7 / Part

10 urves (cont.) The length of a curve is given by Im L b () t dt a A curve is thus a mapping a b Re of the real number line onto the complex plane EE ~ Page 8 / Part

11 Two Special urves ircle The parametric description i Im for a circle centred at complex point a and with a radius r is a r Re it t () are, t[, ] EE ~ Page 9 / Part

12 Two Special urves Straight Line The parametric description i of a straight line segment with starting point a and endpoint b is Im a b Re t () ( ba) ta, t[,] EE ~ Page / Part

13 Example : Parametric Representation and Length of urves a) The line segment that connects the points i3 and 5i6 b t () (7i9) t( i3), t[,] L () t dt 7 i 9 dt 7 9 dt 3 a b) The circle with radius and centre i b t () ( i) e, t[, ] it it it L () t dt ie dt i e dt dt 4 a 7 9 EE ~ Page / Part

14 Example (cont.) c) yx, x3 xt, t3 y Let t t () tit ( ), t3 b 3 3 L () t dt i dt dt a EE ~ Page / Part

15 omplex Functions of a omplex Variable A complex function of a complex variable maps one plane to another plane. D Im[] Im[w] w = f() w Re[] domain Re[w] image f ( ) These functions are of the form f ( xiy) uivu( x, y) iv( x, y). w EE ~ Page 3 / Part

16 omplex Mapping A complex-valued function w f ( ) f( x iy) uxy (, ) ivxy (, ), x iy defines a mapping of a domain, D, onto its image the w-plane. For any point in D, we call the point w f ( ) the image of. Similarly, the points of a curve are mapped onto a curve on the w-plane. Im[] w = f() Im[w] w Re[] Re[w] EE ~ Page 4 / Part D domain image

17 Example: a) w f( ) ( xiy) x y ixy uiv The function f () = would for example map the complex number + i to i and the number i to 3 i 4. x t, t[,], y The line segment x [,], y or is mapped to the line segment, since u t, t [,], v u iv( xiy) ( ti) t EE ~ Page 5 / Part

18 Example: (cont.) In polar coordinate: i i i w f( ) Re ( re ) r e For example, the image of the region r 3/, /6 /3 w under the mapping is R 9/4, /3 /3 v y x 3 u EE ~ Page 6 / Part

19 Example: e b) w f ( ) e e xiy x e (cos y i sin y ) x e cos yie sin y uiv x For example, f ( i) ecosiesin EE ~ Page 7 / Part

20 Example: e (cont.) In terms of polar coordinates Therefore i xiy x iy w f ( ) Re e e e e x, R e y Note that we EE ~ Page 8 / Part

21 Example: e (cont.) For w e, consider the images of: y v. Straight lines and y y const x x const x x e x u From R e, y, we see x x x that is mapped onto the circle w e x and is mapped onto the ray y y y y x v y u arg( w) y x R e, y EE ~ Page 9 / Part

22 Example: e (cont.) D xiy / a xb, c y d. Rectangle { }: x R e, y From (a), we can conclude that t any rectangle with side parallel l to the coordinate axes is mapped onto a region bounded by portions of rays and circles. Therefore the image of D is y i a b D { w Re / e Re, cd} v EE ~ Page 3 / Part d c a b x e b e a c d u

23 Example: e (cont.) 3. The fundamental region : y The fundamental region is mapped onto the entire w-plane, excluding the y origin. The strip is mapped onto the upper half-plane y v x - u More generally, every horiontal strip is mapped onto the full w-plane excluding the origin. c yc x R e, y EE ~ Page 3 / Part

24 Example: ln The natural logarithm w = ln () is the inverse relation of the exponential function e, i.e., It follows ln ln e, e iarg ln iarg ln ln e ln e ln iarg EE ~ Page 3 / Part

25 More Example c) f( ) i i e e i( x iy) i( x iy) ix y ix y e e e e y y e (cos xisin x) e (cos xisin x) y y y y e e e e cos x isin x cos xcosh yisin xsinh y uiv EE ~ Page 33 / Part

26 More Example Alternatively, f () coscos( xiy) cos( () x)cos( iy ) sin( () x)sin( iy ) cos( x)cosh( y) isin( x)sinh( y) uiv since sin( ix) isinh( x), cos( ix) cosh( x), tan( ix) itanh( x). Keep in mind that sin sinh i i i i e e e e cos i e e e e cosh EE ~ Page 34 / Part

27 Observations Identities for trigonometric functions also hold for complex arguments, e.g. cos sin sin( ) sin cos cos sin cos( ) cos cos sin sin sin( ) cos, cos( ) sin Also note that sin sin xcosh y icos xsinh y, cos cos xcosh y isin xsinh y EE ~ Page 35 / Part

28 Observations f () w is continuous at the point x iy if uxy (, ) and vxy (, ) are continuous at x iy. For example, f ( ) is continuous everywhere, because and v xy are continuous everywhere. u x y It can be shown that the regular rules of differentiation and integration are still valid, e.g. d d n n n d sin cos d EE ~ Page 36 / Part

29 omplex Integral onsider the curve : t (), t t[, ] and the complex function f which is continuous on. The complex integral of f along is then defined as f( ) d f ( t) ( t) dt EE ~ Page 37 / Part

30 Example 4 a) Let : ( t) ti3, t t b) Let be a circle with radius r and f ( ) and centred at the origin and f ( ) / d (ti 3 t ) ( i 3) dt : ( t) re it, t[, ] ( i3) ( i 3) i t dt / d ire re it it dt i dt i EE ~ Page 38 / Part

31 Properties of omplex Integrals As for real integrals, the following rules apply:. [ f ( ) g( )] d f( ) d g( ) d. kf( d ) k f( d ), kcomplex EE ~ Page 39 / Part

32 Properties of omplex Integrals f ( ) d ( ) 3. f ( ) d f( ) d f( ) d 4. * f d * EE ~ Page 4 / Part

33 Estimation of a omplex Integral t : t ( ), t [, ] f ( ) Let f () be continuous on. If on, then M f ( ) d ML where L is the length of the curve, ie i.e. dx dy L () t dt dt dt dt EE ~ Page 4 / Part

34 Estimation of omplex Integral An Illustration Graphically, take real integration as an example, M f(t) a b t b a f () tdt L = b a shaded area with red lines M L EE ~ Page 4 / Part

35 Estimation of omplex Integral An Illustration For complex cases, for example, we take f () = 3 and to be a unit circle f ( ) f ( ) 3 3 M f ( ) d ML EE ~ Page 43 / Part

36 Example 5 Find the upper bound for the absolute value of e d where is the line connecting the points i 3 and 4i 6 t ( ) ( i3) t ( i3), t[,] ( t ) i ( 3t 3 ), t[, [ ] x( t) i y( t) i3 4i6 L L 3 dt 3 6 M x iy x iy x iy x e e e e e e e x t 4 M e e e xlargest t () t e Thus, 4 e d ML 3 e t EE ~ Page 44 / Part

37 Example 6 a b a b Show that 4 d 3 6 f() 4 Noting that - -4 ( ) ( ) and thus on the circle =, we have 3 M 4 d ML f ( ) it e L M EE ~ Page 45 / Part t

38 Analytical Functions A function f () defined in domain D is said to be an analytic function if it is differentiable with a continuous first order derivative in all points of D. The function f () is said to be analytic at a point in D if f is analytic in a neighbourhood of. D neighborhood on a complex plane is a disc EE ~ Page 46 / Part

39 Analytical Functions Discrepancy of Definitions of Analytic Functions Definition iti given by most popular texts t (e.g., the reference text): t) A function f () defined in domain D is said to be analytic if it is differentiable at every point of D. Definition given in some odd books or the text by Garg et al.: A function f () defined in domain D is said to be analytic if it is differentiable with a continuous first order derivative at every all point of D. Nevertheless, we will use the definition given by Garg et al. throughout EE ~ Page 47 / Part

40 Singularities Points where a function is not analytic are called singular points or singularities or poles sometimes. Example: f( ) is analytic everywhere in D except =, which is thus the singular point or pole of the function. X D Note that a function is either analytic or singular at any given point EE ~ Page 48 / Part

41 Analytical Functions Observations: It can be shown that t the existence of continuous first derivatives implies the existence of a continuous second derivative, etc. This also implies the existence of a Taylor series. ( ) f ( ) f ( )( ) f ( )! An analytic function can thus also be defined as a function for which a Taylor series expansion exists. EE ~ Page 49 / Part

42 Theorem If a function f () = u + ivis analytic in D, then the following auchy-riemann equations hold, i.e. u v u v and x y y x Alternatively, if the auchy-riemann equations hold for a function f () = u + iv and the function has continuous first order partial derivatives, then f () is analytic in D. EE ~ Page 5 / Part

43 Proof. f ( ) EE ~ Page 5 / Part

44 Example f ( ) x y i xy u iv u x v, u y x y y x v and dthe partial lderivatives are continuous. onsequently, f() is analytic. EE ~ Page 5 / Part

45 Example * x y f ( ) i uiv x y x y u y x v, u xy v ( x y ) ( x y ) x y y x f ( ) is analytic everywhere, except where x y i.e. at the origin. EE ~ Page 53 / Part

46 Example * f ( ) xiy uiv u v u,, v x y y x f() is not analytic anywhere EE ~ Page 54 / Part

47 Example f ( ) x y ix y uiv u v u v xy, 4 x y, x y, 4xy x y y x The auchy-riemann equations only hold for x = and/or y =. Since the function is not analytic in a neighbourhood of x = or y =, f () is not analytic anywhere. y = x = EE ~ Page 55 / Part

48 Observations. The sum or product of analytic functions is analytic.. All polynomials are analytic. 3. A rational function (the quotient of two polynomials) is analytic, except at eroes of the denominator. 4. An analytic function of an analytic function is analytic. e,sin,cos,sinh,cosh 5. Functions are analytic everywhere. EE ~ Page 56 / Part

49 auchy s Integral Theorem A domain D in the complex plane is called simply connected if every closed curve in D only encloses points in D. A domain that is not simply connected is called multiply connected. Simply connected Doubly connected Triply connected EE ~ Page 57 / Part

50 auchy s Integral Theorem If f () is analytic in a simply connected domain D, then for every closed path in D D f ( ) d Fundamental Theorem of alculus If F() is an analytic function with a continuous derivative f () = F'() in a region D containing i a piecewise i smooth (pws) arc : = (t), t () ( ) () f( ) d F( ( )) F( ( )) EE ~ Page 58 / Part

51 Applications of auchy s Theorem Applications:. If f () is analytic in a simply connected domain D, then the integral of f () is * independent of path in D. f ( ) d f( ) d * f ( ) d f ( ) d f ( ) d * EE ~ Page 59 / Part

52 Applications (cont.). onsider a doubly connected domain D. If the function f () is analytic in D, then the integral of f () is the same around any closed path that encircles the opening. * f ( ) d f ( ) d * f ( ) d f ( ) d f ( ) d * EE ~ Page 6 / Part

53 Applications (cont.) 3. The integral along a closed path of the function f () which is analytic in the multiply pyconnected domain D, is given by the sum of the integrals around paths which encircle all openings within the region bounded by, e.g. f ( ) d f ( ) d f ( ) d * * 3 Thus f ( ) d f( ) d f( ) d * * 3 f ( ) d f( ) d 3 * * 3 3 EE ~ Page 6 / Part

54 Applications (cont.) 4. In general, it can be shown that if the path encloses the point, then n n ( ) d i n Proof. Without loss of any generality, we assume =. For n, n is analytic anywhere on the whole complex plane. By auchy s theorem, its integration over any closed curve is. For n =, By Application, the integration over any path enclosed is the same. It was shown earlier that t the integration ti of / over a circle : ( t) re it, t[, ] it ire d dt i dt i it re EE ~ Page 6 / Part

55 Proof of Application 4 (cont.) For n <, let m = n > and integrate over : ( t) re it, t[, ]. We have it ire i i( m) t d dt e dt r e r m m imt m i r im ( ) i ( m ) t e m ( m) r m e i( m) cos ( m) isin ( m) m ( m ) r m ( m) r EE ~ Page 63 / Part

56 Example / (a) with the ellipse d x 4y The equation of the ellipse is given by ( xx ) ( y y ) a b a (x, y ) b EE ~ Page 64 / Part

57 Example (cont.) onsequently, the ellipse x 4y is centred at the origin. The function f () = / is differentiable everywhere except at =. The integral of f () is therefore the same for any path which encloses the origin. may thus be replaced with a circular path of radius, i.e., t () e it, t [ [, ] ½ ie d dt i it e it EE ~ Page 65 / Part

58 Example e 9 d The function f( ) e /( 9) is analytic inside the region enclosed by the curve EE ~ Page 66 / Part

59 Example d i ( i) / / d The function is analytic inside the region enclosed by the path of integration. - ½ EE ~ Page 67 / Part

60 auchy s Integral Formula Let D be a simply connected domain with a fixed point in D. Let f () be analytic in D. f ( ) Then is not analytic in D. f ( ) onsequently, d if is a closed curve enclosing However, the integral f ( ) d will be the same for any path enclosing EE ~ Page 68 / Part

61 auchy s Integral Formula it Re, t[, ] onsider a circle with centre. Then f ( ) d f( Re ) Re it it ire it dt i f ( Re ) dt it Since f () is continuous and the integral will have the same value for all values of R, it follows that it d i f Re dt i f dt if dt if R f ( ) lim ( ) ( ) ( ) ( ) EE ~ Page 69 / Part

62 auchy s Integral Formula This leads to auchy's integral formula, stating the following: Let f () be analytic in D. Let be a closed curve in D which encloses. Then f( ) d i f ( ) EE ~ Page 7 / Part

63 Examples (a) 3 sin d i sin (b) d ( 3) d i i 3 3 i 3 3 EE ~ Page 7 / Part

64 Power Series as Analytic Function A power series is of the form n n c ( ) c c ( ) n onvergence: Every power series c ( ) n has a radius of c ( ) convergence r* such that the series is absolutely convergent for n * and divergent for * r r. Example: The following well known geometric series has an r* = : n EE ~ Page 7 / Part

65 Power Series as Analytic Functions The radius of convergence is r * onvergent given by r * c n lim n c n Divergent Example: EE ~ Page 73 / Part Each power series defines a function which is analytic inside the radius of convergence.

66 Integration and Differentiation of a Power Series Inside the radius of convergence, the power series f ( ) c ( ) n n n can be integrated and differentiated on a term-by-term basis, i.e. f ( ) d c ( ) n d n n d d f ( ) n n * nc ( ), r n EE ~ Page 74 / Part

67 Analytic Functions as Power Series Theorem: Let f () be an analytic function in domain D. Let be a point in D and R be the radius of the largest circle with centre lying inside D. Then there is a power series c ( ) n which converges to f () for R. Furthermore n n c n ( n f ) ( ) f ( ) n! i ( ) n d where is the closed circle which encloses. EE ~ Page 75 / Part

68 Analytic Functions as Power Series D R From the last equation, also note that f ( ) ( ) n d i f ( n ) ( ) n! EE ~ Page 76 / Part

69 Examples sin i d d sin 3! d (a) ii 6 (b) 3 e id d e 4 3 3! d ( ) i 4 e i 3e e 3 3 EE ~ Page 77 / Part

70 Laurent and Taylor Series Expansions of omplex Functions If a function f () is analytic for R, then f () has a Taylor series expansion where R n f ( ) a ( ) n n a n f ( ) i ( ) n d f ( n) ( ) n! EE ~ Page 78 / Part

71 Laurent and Taylor Series Expansions of omplex Functions If a function f () is analytic in theringarea R R, R then f () has a Laurent series expansion R R where f ( ) a ( ) n n n a n f ( ) i ( ) n d Points where a function is not analytic are called singularities. EE ~ Page 79 / Part

72 Example (a) The functions e, sin and cos are analytic functions, and have Taylor series expansions with a centre of 3 e * cn r lim! 3! n! n sin lim n! 3! 5! 7! n ( n )! 4 6 cos! 4! 6! c c n n ( n )! lim n n! lim( n ) n EE ~ Page 8 / Part

73 Example (cont.) e sin (b) The functions and are not analytic in the point =. In the 3 region excluding the point =, these functions have Laurent series expansions of 4 e sin! 3! 3 3! 5! 7! EE ~ Page 8 / Part

74 Example (cont.) (c) The function is analytic for. The Taylor series expansion with centre = of this function is the geometric series, i.e., (d)* Find Taylor series expansion of f( ) at = 3 & its convergence radius ( 3) ( 3) f ( ) The series converges for all 3 3. Thus, its 3 * 3 r 3. EE ~ Page 8 / Part

75 lassification of Singularities Poles onsider the Laurent expansion of different functions: sin. No negative powers of in the expansion. For example has a singularity at =. sin 4 6 3! 5! 7!, so that its Laurent expansion has no negative powers of ( ). The function is said to have a removable singularity at =. EE ~ Page 83 / Part

76 Poles (cont.). A finite number of negative powers of in the expansion, e.g., e 3 3! 3! 4! The highest negative power is 3. This function is said to have a 3rd order pole at =. 3. An infinite number of negative powers of in the expansion, e.g., / e 3! 3! This function is said to have an essential singularity at =. EE ~ Page 84 / Part

77 Example (a) f ( ) cos 4 6! 4! 6! 3 5! 4! 6! The function f () has a removable at =. sin (b) f( ) ! 5! 7! 4 3! 5! 7! The function f () has a 4th order pole at =. EE ~ Page 85 / Part

78 Example (cont.) / (c) f ( ) e 3! 3!! 3! The function f () has an essential singularity at =. (d) f ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Thus, f () has nd order pole at =. EE ~ Page 86 / Part

79 Example (cont.) (e) f ( ) ( ) It is clear that f () has singular points at = & =, respectively. For =, we have the following o Laurent series es of f () centered ed at = f( ) ( ) ( ) 3 ( ) ( ) ( ) 3 Thus, the order of singularity of f () at = is. EE ~ Page 87 / Part

80 Example (cont.) For =, we have the following Laurent series of f () centered at = f( ) ( ) ( ) 3( ) ( ) ( ) ( ) 3 ( ) ( ) ( ) 3 Thus, the order of singularity of f () at = is again equal to. st order poles are also called simple poles or simple singularities. EE ~ Page 88 / Part

81 Zeros If g( ), then the function g () is said to have a ero or root in =. ( n) ( n) If g ( ) g( ) g( ) g ( ) and g ( ), then the function is said to have an nth order ero in =. Theorem: If the function g () has an nth order ero in =, then f( ) has an g ( ) n-th order pole in =. EE ~ Page 89 / Part

82 Example a) onsider the function f( ), which has a singularity at ( )( e e) Let g( ) ( )( e e) Then ( ) g e e( ) e g() g( ) e ( ) e e g() e Therefore g ( ) has a nd order ero in =, and f () has a nd order pole in =. EE ~ Page 9 / Part

83 Example (cont.) f( ) sin b) onsider, which has a singularity at g ( ) sin Let. Then g( ) cos g() g ( ) sin g () g( ) cos g() g ( ) Therefore has a 3rd order ero in =, and f () has a 3rd order pole at =. EE ~ Page 9 / Part

84 Order of Singularities The following method is modified from a suggestion made by Ang Zhi Ping, a student taking EE in Semester of Year 7/8. Given a function f () = h () / g () and g ( ) =, the order of the pole at = can be determined without finding Laurent series as follows:. Find the order (say n) of ero of g () at =.. Find the order (say m) of ero of h () at =, if it is a ero of h (); Otherwise, m =. The order of the pole of f () at = is given by n m. Note that there are m pole-ero cancellations between the numerator and denominator. EE ~ Page 9 / Part

85 Example An alternative solution to Q... in Tutorial., i.e., finding the order of poles for f( ) h ( ) e sin g ( ). It is obvious that g () has a ero of order at =, i.e., n =.. h () e sin Noting h() e cos, we have m =. h e sin 3. The order of the pole of f () at = is given n m =. It is removable. EE ~ Page 93 / Part

86 Residues We know that if f () is analytic in domain D except at the point, it has a Laurent series expansion of f ( ) a ( n n ) n with a n f ( ) i ( ) n d and any closed curve in D which encloses. EE ~ Page 94 / Part

87 Residues From last expression of a n, it follows that a f( ) d Res( f, ) i where Res( f, ) is known as the residue of f at. Thus f ( ) d i Res( f, ) EE ~ Page 95 / Part

88 Residues If f () is not analytic in several points,, n, then f ( ) d f( ) d f ( ) d f( ) d n i Res( f, ) i Res( f, ) i Res( f, ) n n i Res( f, i ) i 3 3 EE ~ Page 96 / Part

89 alculation of Residues. f () has a simple pole at = : Res( f, ) lim ( ) f( ). f () has an nth order pole at = : n d n Res( f, ) lim ( ) f( ) ( n )! n d A( ) 3. f( ) where B( ) has a simple ero at, while B ( ) A ( ) A( and both A and B are differentiable at = : ) Res( f, ) B ( ) EE ~ Page 97 / Part

90 alculation of Residues '. The nd formula of the previous page can be modified as follows: Assuming that f () has an n-th n order pole at =, then m d m Res( f, ) lim ( ) f ( ) m ( m )! d where m is any integer with m n. This formula was proposed and derived by Phang Swee King, a student taking EE in Semester of Year 7/8. It may yield a simpler way in computing the residue for certain situations. EE ~ Page 98 / Part

91 Example using Formula ' Example: The extra freedom in selecting m in Formula ' can simplify some problems in computing residues and thus complex integrals. We consider the following example (which was solved earlier), sin 4 d It was shown that the function has a 3rd order pole at =. However, if we use m = 4 instead of 3, it is much easier to compute the associated residue compared with that using the original formula, i.e., 3 3 sin d 4 sin d i d i lim i lim sin i cos 3! d 3! d 3! 3 EE ~ Page 99 / Part

92 Example 3 (a) alculate 4 3. d Res( f, ) lim ( ) f ( ) f( ) ( ) has simple poles in = and =. 4 3 d i Res( f,) Res( f,) i lim lim i[ 4] 6i EE ~ Page / Part

93 Example 3 (b) ompute e d e d i Res( f,) ilime i Res( f, ) lim( ) f ( ) EE ~ Page / Part

94 Example 3 (c) ompute sin d sin f ( ) has a simple pole at =. Thus sin sin sin d d i lim i Res( f, ) lim( ) f ( ) EE ~ Page / Part

95 Example 3 n d n Res( f, ) lim ( ) f( ) ( n )! n d (d) ompute ( 4)( ) 5 d The function f( ) has a simple ( 4)( ) pole at = 4 and a nd order pole at = Res( f, 4) lim 4 ( ) 5 Res(,) lim d f! d 4 ( 4) 8 lim ( 4) 5 ( 4)( ) 5 d i Res( f, 4) Res( f,) EE ~ Page 3 / Part Res( f, ) lim ( ) f ( )

96 Example 3 (e) ompute e d e d i Res( f,) i e i Res( f, ) A( ) B ( ) EE ~ Page 4 / Part

97 Example 3 4 (f) ompute d f ( ) ( 3)( ) has a simple pole at = enclosed by. Thus, i d i Res( f,) i lim Res( f, ) lim ( ) f ( ) EE ~ Page 5 / Part

98 Applications Real Integral of the form f (cos,sin ) d These integrals can be transformed to an integral of a complex function along the circle. The circle can be described by i e,, Then cos i i e e sin e i e i i i d d d d i ie d EE ~ Page 6 / Part

99 Real Integrals onsequently, we have ie f cos,sin d f, d i i ie i f, d i i EE ~ Page 7 / Part

100 Example 4 (a) Evaluate cos d d d cos i i i d 4 ( 3) ( 3) i i Res( f, 3) d Res( f, ) lim ( ) f ( ) EE ~ Page 8 / Part

101 Example 4 (b) ompute 5 3sin d 5 3sin d d -i/3 s 5 3 i i 3 i 3 d 3 d i 3 d i Res( f, i ) 3 3 i 3 i3 3 -i3 4i lim / 3 i 3 i3 i( ab) ab ( ib)( ia) EE ~ Page 9 / Part

102 Improper Integrals of Rational Functions We now consider real integrals for which the interval of integration is not finite. These are called improper p integrals, and are defined by f ( x) dx lim f ( x) dx R R R Assume that the f( x) P( x) Qx ( ) is a real rational function with Qx ( ) for all real x (i.e. no real poles) degree[ Qx ( )] degree[ Px ( )] EE ~ Page / Part

103 Improper Integrals of Rational Functions (cont.) onsider the complex integral f ( ) d with as indicated in the figure below. Since f (x) is rational, f () will have a finite number of poles in the upper half-plane plane, and if we choose R large enough, encloses all these poles. S -R R Note that consists of a straight path from R to R and a half circle S on the upper plane. EE ~ Page / Part

104 Improper Integrals S -R R Then R f ( ) d f( ) d f ( ) d (*) R S The nd condition, i.e., degree [Q(x)] degree [P(x)] +, implies if n n m m n n m m Q ( ) q q q, P ( ) p p p then, n m ++. Thus, for on S when R is large m m P ( ) pm pm pm K K K ( ) ( ) n n n m Q ( ) qn qn qn R f K f K K f ( ) d ML R as R (Result S) R R S EE ~ Page / Part

105 Improper Integrals of Rational Functions (cont.) And consequently lim f ( ) d. From the equation (*) on last slide, R S we therefore have thatt or R R R lim f ( ) d f ( ) d S f ( xdx ) i Res( fa, j ) -R R j where the sum is taken over all the poles in the upper half-plane. EE ~ Page 3 / Part

106 Example 5 (a) alculate l Let x dx f( ) ( i)( i) +i x dx i Res( f, i) i -i i Res( f, ) lim ( ) f ( ) EE ~ Page 4 / Part

107 Example 5 (b) alculate 3 ( x ) dx ( ) +i Let f( ) ( ) ( i ) ( i ) Res(, ) 3 ( ) dx x i f i ilim i 5 ( i) 6i 3 i 6 8 -i n d n Res( f, ) lim ( ) f ( ) ( )! n n d EE ~ Page 5 / Part

108 Example 5 (c) alculate 4 x 4 dx. Let f( ) 4 and its poles are given by 4 i(n) 4 4 i(n) 4 4 i (n ) e 4 e e, i() i 4 4 n,,, 3, e e i, i( ) i3 4 4 i ( ) i5 4 4 n e e i, n e e i, 3 i( 3 ) i7 4 4 n 4 e e i, n 3 -+i +i EE ~ Page 6 / Part

109 Example 5 Then 4 x 4 dx i Res( f, i) Res( f, i) i 4 4 i ( i) ( i) 6 6 i i i i -+i +i Res( f, ) A( ) B ( ) EE ~ Page 7 / Part

110 Improper Integrals of Fourier-type onsider integrals of the form f x cos mxdx or f xsin mxdx Assume that f (x) = P(x) / Q(x) is a real rational function with Q(x) for all real x (i.e. no real poles) degree [Q(x)] degree [P(x)] + m > EE ~ Page 8 / Part

111 Improper Integrals of Fourier-type onsider the complex integral R im im im im f ( ) e d f ( ) e d f ( ) e d i Res( f e, a j ) R S We will show (i.e., Theorem X next page) that under the conditions m > and im degree [Q(x)] degree [P(x)] +, we have f ( e ) d as R. Noting im that e cos misin m, it can be shown S j im f ( x)cosmxdxrei Res( f e, aj ) j S im f ( x)sinmxdximi Res( f e, aj ) j -R R EE ~ Page 9 / Part

112 Theorem X P( ) If g ( ) with degree[ Qx ( )] degree[ Px ( )] and m >, then Q ( ) im lim ge ( ) d R S S -R R There is no name for this theorem. As such, for easy references, we call it Theorem X. The result has been used earlier in deriving improper integrals of Fourier-type. It will be used later few more times. EE ~ Page / Part

113 Proof of Theorem X (self-study) study) Observing the curves on the right, if we let X (( Y and R ), the integral of the function along S is the same as it along the blue straight lines. Under that degree [Q(x)] degree [P(x)] + iy S -R X X R and along the straight line from X + i to X + iy, we have K K K g( ) K g( ), X iy X and thus X iy Y im im( X iy) imx my my e e e e e im K my K my K ge ( ) d e dy e as X X X m mx X i EE ~ Page / Part

114 Proof of Theorem X (cont.) (self-study) study) Similarly, we can show the integral along the straight line from X + iy to X + i has a same bound, i.e., iy S Xi X iy im K ge ( ) d as X mx The integral along the line from X + iy to X + iy, we have K K K g( ) K g( ), x iy Y and thus XiY my X -R X X R im im( xiy ) imx my my e e e e e im Ke X my ge ( ) d dx K e as X, Y Y Y X iy X QED EE ~ Page / Part

115 Example 6 (a) i xcos x e dx Re i Res, i x +i Re i i Re e i e i -i Res( f, ) A( ) B ( ) EE ~ Page 3 / Part

116 Example 6 i xsin x e (b) dx Im i Res, i x +i e Im i i Im e / e i i -i Res( f, ) A( ) B ( ) EE ~ Page 4 / Part

117 Example 6 (c) i cos x e dx Re i Res,3i 9x 9 +i 3i e Re i i 3i e e Re i 3i Res( f, ) A( ) B ( ) EE ~ Page 5 / Part

118 Example 6 -+i +i i sin x e dx Im i Res, i x x (d) i i e e e Im i Im i i i i e (cosisin) Im Im e e sin e Res( f, ) A( ) B ( ) EE ~ Page 6 / Part

119 Simple Poles on the Real Axis Up to now, we have only considered functions which do not have poles on the real axis. We can avoid this pole by sidestepping it with a small half-circle of which the radius tends to ero. onsider a function f () which has a simple pole at the point = a on the real axis. The Laurent series of this function is then given by f ( ) a ( a) n n n a g ( ) a x () x x x x x S where the function g () is differentiable in the R a R neighbourhood of the point = a. EE ~ Page 7 / Part

120 Simple Poles on the Real Axis Let ( ) be the circle segment with parametric description i ( ): a e, Then f ( ) d ( ) ( ) a d a gd ( ) ( ) () i ie a d g( ) d ia i e ( ) ( ) g( ) d a EE ~ Page 8 / Part

121 Simple Poles on the Real Axis Since g () is differentiable, it is continuous and bounded, and therefore g( ) d ML ( ) M ( ) () This results in Jordan's lemma, stating that a lim f ( ) d ( ) ia ires( f, a) EE ~ Page 9 / Part

122 Example 7 (a) alculate sin x dx x i onsider, with the path as indicated in the illustration. -R - e () d S R e Because the function f( ) is analytic in the domain enclosed by, and according to auchy's integral theorem, the integral must be ero, i.e., i e d R ( ) S i R i e d EE ~ Page 3 / Part

123 Example 7 Let and R. Then, by Jordan s lemma lim f ( ) d i Res( f,)= ilime i i ( ) lim f( ) d By Theorem X, we have. R S i e d R ( ) S R i e d i i e e d ( i ) d i e d i cos x dx x sin x dx x EE ~ Page 3 / Part

124 Example 7 (b) alculate onsider ( x )( x ) dx () -R i -i S R ( )( ) ( )( ) dx d R ( ) S ( )( ) R i Res( f, i ) EE ~ Page 3 / Part

125 Example 7 Note i i Res( f, i ) i lim i ( )( i ) 3 i lim f ( ) d i Res( f, )= i lim 3 ( ) i S lim f ( ) d (Result S) R S and dx ( )( ) ( )( ) d R R ( ) S -R () ( ) i R i Res( f, i ) i i d dx ( )( ) 3 3 ( x)( x ) 3 EE ~ Page 33 / Part

126 alculation of Inverse Laplace Transforms Laplace transform of a function f (t) is defined as F() s L f() t e f() t dt st Inverse transform: st f () t L F() s e F() s ds i y c x onditions for the existence of an inverse Laplace transform of F(s): lim Fs () and lim sfs () finite s s In terms of the complex variable : t t f( t) e F( ) d Res[ F( ) e, s i ] i i EE ~ Page 34 / Part

127 Example 8 Find the inverse Laplace transform of Solution: Fs () t t e e f ( t) Res, i Res, i s t t e e x c i i it i i it e e sin t i Res( f, ) lim( ) f ( ) y EE ~ Page 35 / Part

128 Argument Principle Argument theorem (the proof can be found in the reference text): Let f () be analytic in a domain D except at a finite number of poles. Let be a simple closed path in D not passing through any of the eroes or poles of f (). Then where f ( ) d n p i f( ) x x o D n = number of eroes of f () inside, counting their multiplicities, p = number of poles of f () inside, counting multiplicities. EE ~ Page 36 / Part

129 Example 9 alculate cot d y cot f ( ) f ( ) Note that with f ( ) sin 3 3 x,,,, Inside, f () has eros of order at and no poles. Then, according to the argument theorem, f ( ) cot d d i(5 ) i f ( ) EE ~ Page 37 / Part

130 Application of Argument Principle Let us focus on the case when f () is a rational function, which is analytic except at possibly a finite number of points. The argument principle p implies f ( ) d d ln f ( ) d ln f ( ) i arg[ f ( )] i f ( ) i i. i dln f( ) d iarg[ f( )] i arg[ f ( )] arg[ f ( )] n p i Note that the ln f () is a real function. EE ~ Page 38 / Part

131 Application of Argument Principle ase : If there is one ero and no pole of f () encircled by on -plane, then arg[ f ( )] ( n p ) ( ) arg[ f ( )] n p i.e., f () will encircle the origin on the image plane once anti-clockwise. f () f ().. f ( ) EE ~ Page 39 / Part f ( )

132 Application of Argument Principle ase : If there is no ero and one pole of f () encircled by on -plane, then arg[ f ( )] ( n p ) ( ) arg[ f ( )] n p i.e., f () will encircle the origin on the image plane once clockwise. f () f () x.. f ( ) EE ~ Page 4 / Part f ( )

133 Application of Argument Principle ase 3: If there is no ero and no pole (or equal numbers of poles and eros) of f () encircled by on -plane, then arg[ f ( )] ( n p ) i.e., f () will not encircle the origin on the image plane at all. arg[ f ( )] n p f () f () EE ~ Page 4 / Part.. f ( ) f ( )

134 Rouche's Theorem Let functions f () and g () be analytic everywhere on and inside. If f ( ) g( ) on, then the total number of eros of p () = f () + g () inside is equal to the number of eros of f () inside. f () f () g () g() EE ~ Page 4 / Part

135 Proof of Rouche's Theorem (self-study) study) Let t [, ]. Since f () and g () are analytic everywhere on and inside and since f ( ) g ( ) on, we have f () + t g () for any (why?). Let f ( ) tg( ) () t d i f( ) tg( ) By Argument Principle (t) = number of eros of f () +tg()inside (why?), i.e., (t) is integer-valued. It can also be shown (pretty complicated!) that (t) is a continuous function of t. Thus, (t) ) can only be a constant, which implies that the number of eros of f () + t g () inside is constant for all t [, ]. The result of Rouche s theorem follows by letting t = and t =, respectively. EE ~ Page 43 / Part

136 Example Determine the number of roots (eros) of that lie within the circle :. 9 6 p( ) 8 hoose 9 6 f( ) 8, g( ). Then on the circle : f( ) g ( ) 6 f ( ) 8 6 g ( ) Thus, on and f () and g () are analytic on and within. As f () has only one root inside, p () = f () + g () also has one root inside. EE ~ Page 44 / Part

137 Example (cont.) roots of f () are marked in *; and Indeed only one of fthem is inside the unit - circle EE ~ Page 45 / Part

138 Example Show that all the roots (eros) of p ( ) a...3., where a is unknown, lie within the unit circle, i.e., :, if a <.3. Let f ( ), g( ) a Then on the unit circle: 6 f() g ( ) a a a...3. a.7 Thus, f ( ) g( ) on and f () and g () are analytic on and inside. As f () has 6 roots inside, p () = f () + g () also has its 6 roots inside. EE ~ Page 46 / Part

139 Example (cont.) a =.95 a =. + i EE ~ Page 47 / Part