wi4243AP: Complex Analysis week 5, Friday K. P. Hart Faculty EEMCS - PowerPoint PPT Presentation

Section 5.4: Laurent series Section 5.5: Analytic continuations Section 6.1: Singularities wi4243AP: Complex Analysis week 5, Friday K. P. Hart Faculty EEMCS TU Delft Delft, 3 october, 2014 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Section 5.5: Analytic continuations Section 6.1: Singularities Outline Section 5.4: Laurent series 1 Example: arctan z An old integral An example Section 5.5: Analytic continuations 2 Uniqueness of analytic functions Reflection Principle Section 6.1: Singularities 3 Definition Classification K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Section 5.5: Analytic continuations Section 6.1: Singularities A reminder Theorem Let f : D → C be analytic, let z 0 ∈ D and let R be the distance from z 0 to the complement of D (if D = C then R = ∞ ). Then on the disc { z : | z − z 0 | < R } we have ∞ � a n ( z − z 0 ) n f ( z ) = n =0 where ( ζ − z 0 ) n +1 d ζ = f ( n ) ( z 0 ) 1 f ( ζ ) � a n = 2 π i n ! C and C is any simple closed contour around z 0 lying inside D. K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Main result Theorem Let f : A → C be analytic, where A = { z : R 1 < | z − z 0 | < R 2 } (an annulus). Then for z ∈ A we have ∞ � c k ( z − z 0 ) k f ( z ) = k = −∞ where for k ∈ Z : 1 � f ( ζ ) c k = ( ζ − z 0 ) k +1 d ζ 2 π i C where C is any simple closed contour around z 0 lying inside A, oriented anticlockwise. K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Why? z r z 0 R 1 R 2 Take z inside the annulus. Take r 1 and r 2 with R 1 < r 1 < | z − z 0 | < r 2 < R 2 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Why? z r z 0 Consider the simple closed curve C consisting of the circle C 2 of radius r 2 (anticlockwise), the segment γ (inward), the circle C 1 of radius r 1 (clockwise), the segment γ (outward) K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Why? Apply Cauchy’s formula: 1 � f ( ζ ) f ( z ) = ζ − z d ζ 2 π i C 1 � f ( ζ ) 1 � f ( ζ ) = ζ − z d ζ + ζ − z d ζ 2 π i 2 π i C 2 C 1 because the integrals along γ cancel. As in the case of Taylor series: ∞ 1 f ( ζ ) 1 f ( ζ ) � � � ( ζ − z 0 ) n +1 d ζ × ( z − z 0 ) n ζ − z d ζ = 2 π i 2 π i C 2 C 2 n =0 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Why? On C 1 we rewrite 1 / ( ζ − z ) = − 1 / ( z − ζ ): 1 1 1 ζ − z = − z − ζ = − z − z 0 + z 0 − ζ 1 1 = − · 1 − ζ − z 0 z − z 0 z − z 0 � n ∞ 1 � ζ − z 0 � = − · z − z 0 z − z 0 n =0 The quotient has modulus r 1 / r < 1 on C 1 , so we get uniform convergence on C 1 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Why? We get � n ∞ � ζ − z 0 1 � ζ − z d ζ = − 1 f ( ζ ) � f ( ζ ) � d ζ 2 π i 2 π i z − z 0 z − z 0 C 1 C 1 n =0 ∞ 1 � f ( ζ )( ζ − z 0 ) n d ζ × ( z − z 0 ) − ( n +1) � = − 2 π i C 1 n =0 ∞ 1 f ( ζ ) � � ( ζ − z 0 ) − n d ζ × ( z − z 0 ) − ( n +1) = − 2 π i C 1 n =0 ∞ 1 � f ( ζ ) � ( ζ − z 0 ) − n +1 d ζ × ( z − z 0 ) − n = − 2 π i C 1 n =1 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Why? We add the results: f ( z ) is the sum of ∞ 1 � f ( ζ ) � ( ζ − z 0 ) − n +1 d ζ × ( z − z 0 ) − n − 2 π i C 1 n =1 and ∞ 1 � f ( ζ ) � ( ζ − z 0 ) n +1 d ζ × ( z − z 0 ) n 2 π i C 2 n =0 We reverse the orientation of C 1 to get rid of the minus-sign. The integrands f ( ζ ) / ( ζ − z 0 ) n +1 are analytic on the whole annulus, so we can replace C 2 and C 1 by one and the same simple closed curve. And this gives us the formula we were looking for. K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example arctan z once more We know: � 1 + iz � arctan z = 1 2 i log 1 − iz The following is another branch cut for arctan z (corresponds to positive real axis): i − i K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example arctan z once more Thus arctan z has a branch that is analytic on the annulus { z : 1 < | z |} . What is the Laurent series? 1 First for 1+ z 2 : � n ∞ ∞ ( − 1) n 1 + z 2 = 1 1 1 = 1 � − 1 � � = z 2 1 + 1 z 2 z 2 z 2 n +2 z 2 n =0 n =0 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example arctan z once more Now integrate: � 1 arctan z = 1 + z 2 d z + c ∞ ( − 1) n � � = z 2 n +2 d z + c n =0 � ( − 1) n ∞ � = z 2 n +2 d z + c n =0 ∞ ( − 1) n +1 � = (2 n + 1) z 2 n +1 + c n =0 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example arctan z once more We have, if | z | > 1: ∞ ( − 1) n +1 � arctan z = c + (2 n + 1) z 2 n +1 n =0 Possible values for c ? arctan 1 = π 4 + k π ( k integer) For z = 1 the series sums to − π 4 So, c = π 2 + k π ( k integer) K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example arctan z once more Observe that � 2 n +1 ∞ ( − 1) n +1 ∞ ( − 1) n � 1 = − arctan 1 � � (2 n + 1) z 2 n +1 = − 2 n + 1 z z n =0 n =0 So, in this case we have arctan z = c − arctan 1 z ; plug in z = 1 to get c = π 2 , so for | z | > 1: ∞ ( − 1) n +1 arctan z = π � 2 + (2 n + 1) z 2 n +1 n =0 K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example An old integral Reconsider a problem from week 4 (Monday): � � What is C arctan z d z ? Same as D arctan z d z . i D − i C K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example An old integral Integrate the Laurent series ∞ ( − 1) n +1 arctan z = π � 2 + (2 n + 1) z 2 n +1 n =0 because | 1 z | = 1 r , where r is the radius of D and r > 1 we have uniform convergence, so ∞ ( − 1) n +1 � � π � 1 � arctan z d z = 2 d z + z 2 n +1 d z (2 n + 1) D D D n =0 � 1 = 0 − z d z = − 2 π i D K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Some Laurent series 1 2+ i ( 1 1 1 Consider f ( z ) = ( z − i )( z +2) = z − i − z +2 ). f is analytic on three annuli around 0: { z : | z | < 1 } { z : 1 < | z | < 2 } { z : 2 < | z |} We make the three Laurent series. K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Some Laurent series First annulus: { z : | z | < 1 } . We have ∞ z − i = − 1 1 1 + iz = − 1 1 � ( − iz ) n i i n =0 and ∞ z + 2 = 1 1 1 = 1 − z � n � � 1 + z 2 2 2 2 n =0 Now add. K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Some Laurent series Second annulus: { z : 1 < | z | < 2 } . We have � i � n ∞ ∞ i n z − i = 1 1 1 = 1 � � = 1 − i z n +1 z z z z n =0 n =0 and ∞ z + 2 = 1 1 1 = 1 − z � n � � 1 + z 2 2 2 2 n =0 Now add. K. P. Hart wi4243AP: Complex Analysis

Section 5.4: Laurent series Example: arctan z Section 5.5: Analytic continuations An old integral Section 6.1: Singularities An example Some Laurent series Third annulus: { z : 2 < | z |} . We have � i � n ∞ ∞ i n z − i = 1 1 1 = 1 � � = 1 − i z n +1 z z z z n =0 n =0 and � n ∞ ∞ ( − 2) n � z + 2 = 1 1 1 = 1 − 2 � � = 1 + 2 z n +1 z z z z n =0 n =0 Now add. K. P. Hart wi4243AP: Complex Analysis