Laurents & Taylors series of complex numbers.pptx
jyotidighole2
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Jan 09, 2024
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Laurents & Taylors series of complex numbers
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ICEEM,AURANGABAD TAYLORS AND LAURENTS SERIES Taylorβs Series If π (π§) is analytic inside and on a circle C with centre at point βaβ and radius βRβ then at each point Z inside C, β² β²β² π ( π§ ) = π ( π ) + ( π§ β π ) π ( π ) + ( π§ β π ) 2 π ( π ) + β― 1! 2! (OR) π ! π ( π§ ) = β β π = (π§βπ) π π π (π) This is known as Taylorβs series of π(π§) about π§ = π. Note: 1 Putting π = in the Taylorβs series we get β² β²β² π ( π§ ) = π ( ) + ( π§ β ) π (0) + ( π§ β ) 2 π (0) + β― this series is called Maclaurinβs Series. 1! 1! Note: 2 The Maclaurinβs for some elementary functions are 1) ( 1 β π§ ) β1 = 1 + π§ + π§ 2 + π§ 3 + β―, when | π§ | < 1 2) ( 1 + π§ ) β1 = 1 β π§ + π§ 2 β π§ 3 + β―, when | π§ | < 1 3) ( 1 β π§ ) β2 = 1 + 2π§ + 3π§ 2 + 4π§ 3 + β―, when | π§ | < 1 4) ( 1 + π§ ) β2 = 1 β 2π§ + 3π§ 2 β 4π§ 3 + β―, when | π§ | < 1 1 ! 2 ! 2 5) π π§ = 1 + π§ + π§ + β― when | π§ | < β 1 ! 2 ! 2 6) π π§ = 1 β π§ + π§ + β― when | π§ | < β 3 ! 5 ! 3 5 7) sin π§ = π§ β π§ + π§ + β― when | π§ | < β 2 ! 4 ! 2 4 8) c o s π§ = 1 β π§ + π§ + β― w h e n | π§ | < β LAURENTS SERIES If π 1 πππ π 2 are two concentric circles with centre at π§ = π and radii π 1 πππ π 2 ( π 1 < π 2 ) and if π ( π§ ) is analytic inside on the circles and within the annulus between π 1 πππ π 2 then for any z in the annulus, we have π ( π§ ) = β β π π ( π§ β π ) π + β β π ( π§ β π ) β π β¦ (1 ) π=0 π=1 π π Where π = π 1 π(π§) β« ππ§ and π = 2ππ π 1 (π§βπ) π+1 2ππ π 2 (π§βπ) 1βπ 1 π(π§) β« ππ§ and the integration being taken in positive direction. This series (1) is called Laurent series of π ( π§ ) about the point π§ = π π Example: Expand π ( π ) = ππππ as a Taylorβs series about π = π
. Solution: Complex calculus
Function Value of function at π§ = π 4 π ( π§ ) = πππ π§ π ( π ) = cos ( π ) = 1 4 4 β 2 π β² ( π§ ) = β π π π π§ π β² ( π ) = βπ ππ ( π ) = β 1 4 4 β 2 π β²β² ( π§ ) = βπππ π§ π β²β² ( π ) = βπππ ( π ) = β 1 4 4 β 2 π β²β²β² ( π§ ) = π πππ§ π β²β² ( π ) = π ππ ( π ) = 1 4 4 β 2 β² π The Taylor series of π ( π§ ) about π§ = π is π ( π§ ) = π ( π ) + (π§ β π ) π ( 4 ) + (π§ β π ) 2 β²β² π 4 4 4 1 ! 4 2 ! π ( 4 ) + β― β 1 2 β 1 β 2 4 1 ! 4 2 ! πππ π§ = 1 + (π§ β π ) β 2 + (π§ β π ) β 2 + β― Example: Expand π ( π ) = πππ (π + π) as a Taylorβs series about π = π . Solution: Function Value of function at π§ = π ( π§ ) = πππ (1 + π§) π ( ) = πππ (1 + 0) = π β² ( π§ ) = 1 1 + π§ π β² ( ) = 1 1 + = 1 π β² β² ( π§ ) = β1 π β²β² ( ) = β1 = β1 ( 1 + π§ ) 2 ( 1 + ) 2 π β²β² β² ( π§ ) = 2 π β²β² β² (0) = 2 = 2 ( 1 + π§ ) 3 ( 1 + ) 3 The Taylor series of π ( π§ ) about π§ = is β² β²β² π ( π§ ) = π ( ) + ( π§ β ) π (0 ) + ( π§ β ) 2 π (0 ) + β― 1! 2! πππ (1 + π§) = + ( π§ ) 1 + ( π§ ) 2 β1 + β― 1! 2! πππ (1 + π§) = ( π§ ) 1 β ( π§ ) 2 1 + β― 1! 2! Example: Expand π ( π ) = π π βπ ( π +π )( π+ π) as a Laurentβs series if (i) | π | < 2 (ii) | π | > 3 (iii) π < | π | < 3 Solution: ICEEM,AURANGABAD Complex calculus
Given π ( π§ ) = π§ 2 β1 (π§+2)(π§+3) is an improper fraction. Since degree of numerator and degree of denominator of π ( π§ ) are same β΄ Apply division process 1 π§ 2 + 5π§ + 6 π§ 2 β 1 π§ 2 + 5π§ + 6 β5π§ β 7 β΄ π§ 2 β1 (π§+2)(π§+3) = 1 β 5 π§ + 7 ( π§ + 2 ) ( π§ + 3 ) β¦ ( 1) C o n s i d e r = + 5 π§ + 7 π΄ π΅ ( π§ + 2)( π§ + 3 ) π§ + 2 π§ + 3 β 5π§ + 7 = π΄ ( π§ + 3 ) + π΅ ( π§ + 2 ) Put π§ = β2, we get β10 + 7 = π΄ (1) β π΄ = β3 Put π§ = β3, we get β15 + 7 = π΅(β1) β π΅ = 8 β΄ 5 π§ + 7 = β3 + 8 ( π§ + 2)( π§ + 3 ) π§ + 2 π§ + 3 β΄ ( 1 ) β 1 β β 3 8 π§ + 2 π§ + 3 (i) Given | π§ | < 2 β΄ π ( π§ ) = 1 + β 3 8 2(1+ π§ β 2 ) 3(1+ π§ β 3 ) 3 ( π§ β 1 8 π§ = 1 + 1 + β ) β (1 + β ) 2 2 3 3 β 1 3 2 π§ 2 2 π§ 2 8 3 π§ 3 3 π§ 2 = 1 + [ 1 β + [ ] + β― ] β [ 1 β + [ ] + β― ] 2 π = 1 + 3 β β π = ( β1 ) [ π§ π 8 2 3 ] β β β π = 3 π ( β1 ) π [ π§ ] (ii) Given | π§ | > 3 β΄ π ( π§ ) = 1 + β 3 8 π§(1+ 2 β π§ ) π§(1+ 3 β π§ ) 3 ( 2 β 8 3 β = 1 + 1 + π§ ) β (1 + π§ ) π§ π§ β 1 β 1 π§ π§ π§ 2 2 3 2 8 π§ 3 π§ π§ 3 2 = 1 + [ 1 β + [ ] + β― ] β [ 1 β + [ ] β¦ ] π π = 2 π§ π§ π = 1 + 3 β β ( β1 ) [ ] β 8 π§ β β π = π§ π ( β1 ) π [ 3 ] (iii) Given 2 < | π§ | < 3 ICEEM,AURANGABAD Complex calculus
β΄ π ( π§ ) = 1 + 3 π§(1+ 2 β π§ ) β 8 3(1+ π§ β 3 ) 3 ( β 1 8 2 β π§ = 1 + 1 + π§ ) β (1 + β ) π§ 3 3 β 1 2 2 = 1 + 3 [ 1 β 2 + [ 2 ] + β― ] β 8 [ 1 β π§ + [ π§ ] β¦ ] π§ π§ π§ 3 3 3 π π = 2 π§ π§ = 1 + 3 β β ( β1 ) [ ] π β 8 3 β β π = 3 π ( β1 ) π [ π§ ] Example: Find the Laurentβs series expansion of π ( π ) = π π β π π(πβπ)(π+π) in π < | π + π | < 3 . Also find the residue of π ( π ) at π = β π Solution: Given π ( π§ ) = 7 π§ β 2 π§(π§β2)(π§+1) = π΄ + + 7 π§ β 2 π΅ πΆ π§ ( π§ β 2 )( π§ + 1 ) π§ π§ β 2 π§ + 1 7π§ β 2 = π΄ ( π§ β 2 )( π§ + 1 ) + π΅π§ ( π§ + 1 ) + πΆπ§ ( π§ β 2 ) Put π§ = 2, we get 14 β 2 = π΅ ( 2 ) (2 + 1) β 12 = 6π΅ β π΅ = 2 Put π§ = β1, we get β7 β 2 = πΆ(β1) ( β1 β 2 ) β β9 = 3πΆ β πΆ = β3 Put π§ = we get β2 = π΄ ( β2 ) β π΄ = 1 β΄ π ( π§ ) = 1 + β 2 3 π§ π§ β 2 π§ + 1 Given region is 1 < | π§ + 1 | < 3 Let π’ = π§ + 1 β π§ = π’ β 1 (π. π) 1 < | π’ | < 3 1 N o w π ( π§ ) = + 2 β 3 π’ β 1 π’ β 3 π’ 1 π’(1β 1 β π’ ) = + 2 β 3 β3(1β π’ β 3 ) π’ 1 ( 1 β β 1 2 π’ β 1 3 = 1 β π’ ) β (1 β β ) β π’ 3 3 π’ π’ π’ π’ 3 1 1 2 π’ 3 3 1 2 π’ 2 = [1 + + [ ] + β― ] β [ 1 + + [ ] + β― ] β 3 π’ 1 π§ + 1 1 [ 1 π§ + 1 π§ + 1 2 2 3 3 3 π§+1 π§+1 2 = [1 + + ] + β― ] β [ 1 + + [ ] + β― ] β 3 π§+1 ICEEM,AURANGABAD Complex calculus
= 1 π§+1 π β β π=0 π§+1 3 π§ + 1 3 [ 1 ] β 2 β β π π = [ 1 ] β 3 π§+1 1 π§ + 1 Also π
ππ [ π ( π§ ) , π§ = β1 ] = coefficient of = β2 Example: Expand π ( π ) = π (πβπ)(πβπ) in a Laurentβs series valid in the region (i) | π β π | > 1 (ii) π < | π β π | < 1 (iii) | π | > 2 (iv) π < | π β π | < 1 Solution: Given π ( π§ ) = π (πβπ)(πβπ) C o n s i d e r = + 1 π΄ π΅ ( z β 1)( z β 2 ) π§ β 1 π§ β 2 β 1 = π΄ ( π§ β 2 ) + π΅ ( π§ β 1 ) Put π§ = 2, we get 1 = π΅(1) β π΅ = 1 Put π§ = 1 we get 1 = π΄ ( 1 β 2 ) β π΄ = β 1 β΄ π ( π§ ) = β1 + 1 π§ β 1 π§ β 2 (i) Given region is | π§ β 1 | > 1 Let π’ = π§ β 1 β π§ = π’ + 1 ( π . π ) | π’ | > 1 Now π ( π§ ) = β 1 + 1 π’ π’ β 1 = β1 + 1 π’ π’(1β 1 β π’ ) β1 1 ( 1 β = π’ + π’ 1 β π’ ) β 1 β 1 1 π’ π’ 1 π’ π’ 1 2 = + [1 + + [ ] + β― ] β 1 1 π§ + 1 π§ + 1 = + [1 + + 1 [ 1 π§ + 1 π§ + 1 2 ] + β― ] β 1 1 [ 1 π§ + 1 π ] β β π = = + π§+1 π§+1 (ii) Given < | π§ β 2 | < 1 Let π’ = π§ β 2 β π§ = π’ + 2 (π. π) < | π’ | < 1 Now π ( π§ ) = β 1 + 1 π’ + 1 π’ = β ( 1 + π’ ) β1 + 1 π’ ICEEM,AURANGABAD Complex calculus
= β [ 1 β π’ + [ π’ ] 2 + β― ] + 1 π’ = β [ 1 β ( π§ β 2 ) + [ π§ β 2 ] 2 + β― ] + 1 π§β2 1 π§β2 = β β β [ β1 ] π [ π§ β 2 ] π + π = (iii) Given | π§ | > 2 Now π ( π§ ) = β + 1 1 1 2 π§(1β π§ ) π§(1β π§ ) β1 β1 = β 1 (1 β 1 ) + 1 (1 β 2 ) π§ π§ π§ π§ 1 2 1 1 1 π§ π§ π§ π§ 2 π§ π§ 2 2 = β [ 1 + + [ ] + β― ] + [ 1 + + [ ] + β― ] π§ π§ π π = β 1 β β [ 1 ] + 1 β ββ [ 2 ] π=0 π§ π=0 π§ (iv) Given < | π§ β 1 | < 1 Let π’ = π§ β 1 β π§ = π’ + 1 ( π . π ) < | π’ | < 1 Now π ( π§ ) = β 1 + 1 π’ π’ β 1 = β 1 + 1 π’ β 1 [ 1 β π’ ] π’ = β 1 β ( 1 β π’ ) β1 π’ = β 1 β [ 1 + π’ + [ π’ ] 2 + β― ] = β β [ 1 + π§ β 1 + [ π§ β 1 ] 2 + β― ] = β 1 π§β1 1 π§β1 β β β [ π§ β 1 ] π π = Example: Expand π ( π ) = π ( π +π )( πβ π) in a Laurentβs series about (i) π = βπ (ii) π = π Solution: C o n s i d e r = + π π¨ π© ( π+ π )( π β π) π + π πβ π β π§ = π΄ ( π§ β 2 ) + π΅ ( π§ + 1 ) Put π§ = 2, we get 2 = π΅(3) β π΅ = 2 3 Put π§ = β 1 we get β1 = π΄ ( β3 ) β π΄ = 1 3 β΄ π ( π§ ) = + 1 2 3 ( π§ + 1 ) 3 ( π§ β 2 ) (i)To expand π(π§) about π§ = β1 ICEEM,AURANGABAD Complex calculus
ICEEM,AURANGABAD (or) | π§ β 1 | < 1 Put π§ + 1 = π’ β π§ = π’ β 1 β | π§ β 1 | < 1 β | π’ | < 1 Now π ( π§ ) = 1 + 3π’ 3(π’β3) 2 = 3 π’ 1 + 2 3((β3)(1β π’ β 3 )) π’ = 1 β 2 (1 β β ) 3π’ 9 3 β 1 1 2 3 π’ 9 π’ 3 3 π’ 2 = β [ 1 + + [ ] + β― ] = 1 2 3( π§ + 1 ) 9 β [1 + ( π§ + 1 ) 3 + [ ( π§ + 1 ) 3 2 ] + β― ] 1 = β 3( π§ + 1 ) 9 3 [ ] π 2 β β (π§+1) π = (ii) To expand π(π§) about π§ = 2 (or) | π§ β 2 | < 1 Put π§ β 2 = π’ β π§ = π’ + 2 β | π§ β 2 | < 1 β | π’ | < 1 Now π ( π§ ) = + 1 2 3 ( π’ + 3 ) 3( π’ ) = 1 3((3)(1+ u β 3 )) + 2 3(u) 1 ( u β 1 2 = 1 + β ) + 9 3 3(u) 9 1 u 3 3 u 2 = [1 β + [ ] + β― ] + 2 3(u) 1 9 = [1 β 3 ( z β 2 ) ( z β 2 ) 3 2 + [ ] + β― ] + 2 3 ( z β 2 ) = 9 n ( z β 2 ) n 1 β β n=0 β1 2 ( ) [ ] + 3 3 ( z β 2 ) π(πβπ)(π+π) Example: Expand the Laurentβs series about for π ( π ) = ππ+π in the region π < | π + π | < 3 Solution: C o n s i d e r 6 π§ + 5 π΅ = π΄ + + π π§(π§β2)(π§+1) π§ π§β2 π§+ 1 β 6π§ + 5 = π΄ ( π§ β 2 )( π§ + 1 ) + π΅π§ ( π§ + 1 ) + πΆπ§(π§ β 2) Put π§ = 0, we get 5 = π΄(β2)(1) Complex calculus
β π΄ = β5 2 Put π§ = β 1 we get β11 = πΆ ( β1 ) (β3) β πΆ = β 11 3 Put π§ = 2 we get 17 = π΅ ( 2 ) (3) β π΅ = 17 6 β΄ π ( π§ ) = β5 + β 1 7 11 2π§ 6(π§β2) 3(π§+ 1) Given region 1 < | π§ + 1 | < 3 Put π§ + 1 = π’ β π§ = π’ β 1 (π. π) 1 < | π’ | < 3 N o w π ( π§ ) = + β5 17 β 11 2 ( π’ β 1 ) 6( π’ β 3 ) 3 π’ = β 5 1 π’ 2π’(1β ) + π’ 17 β 11 6(β3)(1β 3 ) 3π’ β5 [ 2 π’ π’ 17 1 8 3 1 β1 π’ β1 = 1 β ] β [ 1 β ] β 11 3 π’ β 5 2 π’ π’ π’ 1 2 1 17 18 π’ 3 3 π’ 2 = [1 + + [ ] + β― ] β [ 1 + + [ ] + β― ] β 11 3 π’ = β 5 2( π§ + 1 ) [1 + 1 1 ( π§ + 1 ) ( π§ + 1 ) 2 17 18 + [ ] + β― ] β [ 1 + 3 3 (π§+1) (π§+1) 2 + [ ] + β― ] β 11 3(π§+1) β 5 = 2( π§ + 1 ) 1 ( π§ + 1 ) π [ ] β 18 11 ] β 3 3( π§ + 1 ) π 17 β β (π§+1) π=0 [ β β π = ICEEM,AURANGABAD Complex calculus
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