EI331 Signals and Systems Lecture 23 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 23 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 16, 2019

Contents 1. Complex Integration 2. Cauchy’s Integral Theorem 3. Cauchy’s Integral Formula 1/22

Integration Recall for f ( t ) = u ( t ) + jv ( t ) of a real variable t , � b � b � b f ( t ) dt = u ( t ) dt + j v ( t ) dt a a a Given a smooth curve γ parameterized by z : [ a , b ] → C , and a function f continuous on γ , the (contour) integral of f along γ is � b � f ( z ) dz � f ( z ( t )) z ′ ( t ) dt γ a � γ f ( z ) dz is independent of reparametrization. Proof. If ˜ z : [ c , d ] → C is a reparametrization of γ obtained from z by ˜ z ( τ ) = z ( t ( τ )) , where t = t ( τ ) is continuously differentiable and t ′ ( τ ) > 0 , then the change-of-variable formula yields � b � d � d f ( z ( t )) z ′ ( t ) dt = f ( z ( t ( τ ))) z ′ ( t ( τ )) t ′ ( τ ) d τ = z ′ ( τ ) d τ f (˜ z ( τ ))˜ a c c 2/22

Integration The integral along a curve is essentially two line integrals of the second kind, which depends on the orientation of the curve, � b � [ u ( x ( t ) , y ( t )) + jv ( x ( t ) , y ( t ))] · [ x ′ ( t ) + jy ′ ( t )] dt f ( z ) dz = γ a � � = ( udx − vdy ) + j ( udy + vdx ) γ γ Let − γ (also γ − ) be the opposite of γ parametrized by z : [ − b , − a ] → C , where ˜ ˜ z ( t ) = z ( − t ) . By a change of variable, � − a � a � � z ′ ( t ) dt = f ( z ( t )) z ′ ( t ) dt = − f ( z ) dz = f (˜ z ( t ))˜ f ( z ) dz − γ − b b γ The positive orientation of a Jordan curve is such that when traveling along it the interior of the curve is always to the left. 3/22

Integration Subdivision of the curve γ If γ is divided into a finite number of segments, denoted by γ = γ 1 + γ 2 + · · · + γ n , then � � � � f ( z ) dz = f ( z ) dz + f ( z ) dz + · · · + f ( z ) dz γ γ 1 γ 2 γ n If γ is piecewise smooth, we simply define the integral along γ to be the sum of the integrals along each smooth segment. NB. We only consider piecewise smooth curves in this course. Linearity in the integrand For a , b ∈ C , � � � [ af ( z ) + bg ( z )] dz = a f ( z ) dz + b g ( z ) dz γ γ γ 4/22

Integration � � � � � � If | f ( z ) | ≤ M on γ , then f ( z ) dz � ≤ | f ( z ) | ds ≤ ML γ , where ds � � � γ γ is the infinitesimal arc length, and L γ is the length of γ . Proof. Let z : [ a , b ] → C be a parametrization of γ . � b � b � � � � � � � � f ( z ( t )) z ′ ( t ) dt � | f ( z ( t )) | · | z ′ ( t ) | dt f ( z ) dz � = � ≤ � � � � � � γ a a | x ′ ( t ) | 2 + | y ′ ( t ) | 2 dt = | z ′ ( t ) | dt , so � Recall ds = � b � | f ( z ( t )) | · | z ′ ( t ) | dt = | f ( z ) | ds . a γ Since | f ( z ) | ≤ M on γ , � b � b � | f ( z ( t )) | · | z ′ ( t ) | dt ≤ M | z ′ ( t ) | dt = M ds = ML γ . a a γ 5/22

Examples Example. Let z : [ a , b ] → C be a parameterization of a smooth curve γ . � b � z ′ ( t ) dt = z ( t ) | b dz = t = a = z ( b ) − z ( a ) γ a � b � z ( t ) z ′ ( t ) dt = 1 t = a = 1 2 z 2 ( b ) − 1 b � 2 z 2 ( t ) 2 z 2 ( a ) zdz = � � γ a NB. The values of the integrals depend only on the endpoints. Example. Let C = { z ∈ C : | z − z 0 | = R } be a circle parameterized by z ( t ) = z 0 + Re jt , t ∈ [ 0 , 2 π ] . For n ∈ Z , � 2 π � 2 π Rje jt � dz ( Re jt ) n dt = jR 1 − n e j ( n − 1 ) t dt = j 2 πδ [ n − 1 ] ( z − z 0 ) n = C 0 0 NB. The value is independent of z 0 and R . 6/22

Example Let γ be the boundary of the annulus y { z : 1 ≤ | z | ≤ 2 } in the first quadrant with 2 j γ 2 positive orientation. j � 2 � 1 γ 1 � � � dz dx zdz jdy zdz z = x + | z | 2 + − jy + ¯ | z | 2 γ 1 γ 2 2 γ 1 x O 1 2 = log 2 + 1 � � zdz + log 2 + zdz 4 γ 2 γ 1 = log 2 + 1 8 ( 4 j 2 − 4 ) + log 2 + 1 ( ∗ ) 2 ( 1 − j 2 ) = 2 log 2 In ( ∗ ) , we have used the first example on the previous slide. 7/22

Example γ 1 has parameterization z 1 ( t ) = ( 1 + j ) t , t ∈ [ 0 , 1 ] y � 1 � 1 � z 0 = 1 + j zdz = ¯ ( 1 − j ) t ( 1 + j ) dt = 2 tdt = 1 γ 1 γ 1 0 0 γ 3 γ 2 has parameterization z 2 ( t ) = t , t ∈ [ 0 , 1 ] γ 3 has parameterization z 3 ( t ) = 1 + jt , t ∈ [ 0 , 1 ] x O γ 2 1 � � � ¯ ¯ ¯ zdz = zdz + zdz γ 2 + γ 3 γ 2 γ 3 � 1 � 1 = tdt + ( 1 − jt ) jdt 0 0 � 1 � = 1 2 + 2 + j = 1 + j The value of the integral depends not only on the endpoints but also on the path. 8/22

Contents 1. Complex Integration 2. Cauchy’s Integral Theorem 3. Cauchy’s Integral Formula 9/22

Cauchy’s Integral Theorem Suppose f is analytic in a simply connected domain D and γ a piecewise smooth Jordan curve in D . Recall � � � f ( z ) dz = ( udx − vdy ) + j ( udy + vdx ) γ γ γ If u , v are continuously differentiable on D , then Green’s Theorem and the Cauchy-Riemann equations imply � � ( udx − vdy ) = ( − v x − u y ) dxdy = 0 Ω γ � � ( udy + vdx ) = ( u x − v y ) dxdy = 0 γ Ω where Ω is the region bounded by γ � Cauchy’s Integral Theorem asserts that γ f ( z ) dz = 0 without explicitly assuming continuous differentiability of u and v . 10/22

Cauchy’s Integral Theorem Theorem. If f ( z ) is analytic in a simply connected domain D , and γ is a D piecewise smooth closed (possibly not simple) curve in D , then γ � � f ( z ) dz = f ( z ) dz = 0 . γ γ Theorem. Let D be the interior of a piecewise smooth Jordan curve γ and D ¯ D = D ∪ γ its closure. If f ( z ) is analytic on D and continuous on ¯ D , then γ = ∂ D � � f ( z ) dz = f ( z ) dz = 0 . γ γ 11/22

Cauchy’s Integral Theorem Theorem. Let γ 0 , γ 1 , γ 2 , . . . , γ n be n + 1 positively oriented piecewise smooth Jordan curves such that (a) γ 1 , . . . , γ n lie in the interior of γ 0 (b) each of γ 1 , . . . , γ n lies in the exteriors of the others Let D be the multiply connected domain with boundary γ 0 , γ 1 , . . . , γ n . If f ( z ) is analytic on D and continuous on ¯ D , then � � � � f ( z ) dz = f ( z ) dz + f ( z ) dz + · · · + f ( z ) dz γ γ 1 γ 2 γ n D − γ 2 − γ 1 γ 12/22

Example � dz Compute z − a , where γ is a piecewise γ smooth Jordan curve and a / ∈ γ . a γ 1 1. z − a is analytic on C \ { a } 2. If a is in the exterior of γ , then � dz z − a = 0 γ 3. If a is in the interior of γ , pick a small γ enough circle γ 1 centered at a that lies γ 1 in the interior of γ a � � 1 1 z − adz = z − adz = j 2 π γ γ 1 13/22

Example Let f ( z ) = 2 z − 1 y z 2 − z and γ be any piecewise γ smooth Jordan curve containing | z | = 1 . � � � 1 1 C 1 C 0 1. f ( z ) dz = z − 1 dz + z dz γ γ γ x 2. 0 , 1 lie in the interior of γ � � 1 1 3. z − 1 dz = j 2 π , z dz = j 2 π γ γ � f ( z ) dz = j 4 π 4. γ If a , b / ∈ γ , int γ and ext γ are the interior and exterior of γ ,  0 , if a , b ∈ int γ or a , b ∈ ext γ  � 1 dz  1 ( z − a )( z − b ) = a − b , if a ∈ int γ , b ∈ ext γ j 2 π γ  1 b − a , if b ∈ int γ , a ∈ ext γ  14/22

Path Independence Theorem. If f ( z ) is analytic on a simply connected domain D , then for any piecewise smooth curve γ in D , the integral � γ f ( z ) dz depends only on the endpoints of γ . Proof. Let z 0 , z 1 ∈ D and γ 1 , γ 2 two piecewise smooth curves in D . Then γ = γ 1 + γ − 2 is a closed curve in D . By Cauchy’s Theorem, � � � f ( z ) dz + f ( z ) dz = f ( z ) = 0 γ 1 γ − γ 2 so � z 1 � � � f ( z ) dz � f ( z ) dz = − f ( z ) dz = f ( z ) dz γ 1 γ − γ 2 z 0 2 If we fix z 0 and let z 1 vary, we can define a function � z F ( z ) = f ( ζ ) d ζ z 0 15/22

Primitive Theorem. If f ( z ) is analytic on a simply connected domain D � z and z 0 ∈ D , then the function F ( z ) = z 0 f ( ζ ) d ζ is analytic on D and F ′ ( z ) = f ( z ) . NB. As in calculus, a function F satisfying F ′ ( z ) = f ( z ) is called a primitive of f . � z +∆ z Proof. Fix an arbitrary z ∈ D . When evaluating f ( ζ ) d ζ , we z 0 can pick a path z 0 → z → z + ∆ z . Then � z +∆ z � z � z +∆ z F ( z + ∆ z ) − F ( z ) = f ( ζ ) d ζ + f ( ζ ) d ζ = f ( ζ ) d ζ z 0 z 0 z Since f is analytic, it is also continuous. Given any ǫ > 0 , there is a δ > 0 s.t. | f ( ζ ) − f ( z ) | < ǫ when | ζ − z | < δ , and B ( z , δ ) ⊂ D . When | ∆ z | < δ , the line segment connecting z and z + ∆ z is contained in B ( z , δ ) . 16/22

Primitive Proof (cont’d). Note � z +∆ z f ( z )∆ z = f ( z ) d ζ z Thus � z +∆ z F ( z + ∆ z ) − F ( z ) − f ( z ) = 1 [ f ( ζ ) − f ( z )] d ζ ∆ z ∆ z z and � z +∆ z � F ( z + ∆ z ) − F ( z ) � 1 � � − f ( z ) � ≤ | f ( ζ ) − f ( z ) | ds ≤ ǫ � � ∆ z | ∆ z | � z This shows F ′ ( z ) = f ( z ) . Since z is arbitrary, F is analytic on D . 17/22