Hence Res R 2 i f 2 i 2 i 1 2 i 2 i 2 i 1 3 24 7 i 500 Res R 2 i g 2 i 2 i 1 2

# Hence res r 2 i f 2 i 2 i 1 2 i 2 i 2 i 1 3 24 7 i

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Hence Res( R ; 2 i ) = f (2 i ) = 2 i + 1 (2 i + 2 i )(2 i - 1) 3 = 24 - 7 i 500 ; Res( R ; - 2 i ) = g ( - 2 i ) = - 2 i + 1 ( - 2 i - 2 i )( - 2 i - 1) 3 = 24 + 7 i 500 ; Res( R ; 1) = h 00 (1) 2! = - 12 125 . (4) Suppose that F and G are analytic in the disc {| z - z 0 | < r } with G ( z 0 ) = 0 but G 0 ( z 0 ) 6 = 0 . Then Res F G ; z 0 = F ( z 0 ) G 0 ( z 0 ) . To see this, write G ( z ) = ( z - z 0 ) g ( z ), where g is analytic in the disc and g ( z 0 ) = G 0 ( z 0 ) 6 = 0 . Hence Res F G ; z 0 = Res F ( z ) ( z - z 0 ) g ( z ) ; z 0 = F ( z 0 ) g ( z 0 ) = F ( z 0 ) G 0 ( z 0 ) . (5) Find the Laurent series at z 0 = 0 for f ( z ) = (sin z ) /z 3 . Solution. f ( z ) = sin z z 3 = z - z 3 3! + z 5 5! - · · · z 3 = 1 z 2 - 1 3! + z 2 5! - · · · , z 6 = 0 . This is the Laurent series of f ( z ) and the principal part is 1 z 2 and coeffient b 1 of term 1 z is zero and hence the residue of f at z 0 = 0 is zero. (6) The Laurent series of e 1 /z can be easily obtained by the power series expansion of e z as follows: e 1 /z = X n =0 1 n ! (1 /z ) n = X n =0 1 n ! z - n = 1 + 1 z + 1 2! z 2 + 1 3! z 3 + · · · . Therefore Res( e 1 /z ; 0) = 1 . Note that e 1 /z has an essential singularity at z 0 = 0 since the Laurent series has infinitely many negative powers. (7) (Exercise #12.) Find the Laurent series of 1 e z - 1 at z 0 = 0 with first four terms. 24 2. Basic Properties of Analytic Functions Solution. Since e z = 1 + z + z 2 2! + z 3 3! + z 4 4! + · · · , we have 1 e z - 1 = 1 z + z 2 2! + z 3 3! + z 4 4! + · · · = 1 z 1 1 + z 2! + z 2 3! + z 3 4! + · · · = 1 z ( a 0 + a 1 z + a 2 z 2 + a 3 z 3 + · · · ) , where 1 1 + z 2! + z 2 3! + z 3 4! + · · · = a 0 + a 1 z + a 2 z 2 + a 3 z 3 + · · · . We can determine a 0 , a 1 , a 2 , a 3 , · · · by multiplication of power series 1 + z 2 + z 2 6 + z 3 24 + · · · ( a 0 + a 1 z + a 2 z 2 + a 3 z 3 + · · · ) = 1 . For example, this equality becomes a 0 + a 1 + a 0 2 z + a 2 + a 1 2 + a 0 6 z 2 + a 3 + a 2 2 + a 1 6 + a 0 24 z 3 + · · · = 1 . Hence a 0 = 1 , a 1 = - 1 2 , a 2 = 1 12 , a 3 = 0 . Therefore, 1 e z - 1 = 1 - 1 12 z + 1 12 z 2 + · · · z = 1 z - 1 12 + z 12 + · · · . This is the Laurent series for 1 e z - 1 at z 0 = 0 . Easily we see the residue of the function at the pole is 1. Exercises. Page 150. Problems 1, 3, 5–8, 11, 14, 15, 22(a, b). 2.6. The Residue Theorem and Its Applications Theorem 2.16 ( The Residue Theorem ) . Suppose f is analytic in a simply-connected domain D except for a finite number of isolated singularities at points z 1 , z 2 , · · · , z N of D . Let γ be a piecewise smooth positively oriented simple closed curve in D that does not pass through any of the points z 1 , z 2 , · · · , z N . Then (2.11) Z γ f ( z ) dz = 2 πi X all z k inside γ Res( f ; z k ) , where the sum is taken over all those singularities z k lying inside γ. Proof. Let Ω be the inside of γ and z 0 1 , z 0 2 , · · · , z 0 n are the isolated singularities that lie in Ω. Let Δ j be the closed disc with center z 0 j that is contained in Ω and be disjoint for j = 1 , · · · , n. Let U be the domain Ω \ ∪ n j =1 Δ j . Then the positively oriented boundary of U is ∂U = γ ( - Δ 1 ) ∪ · · · ∪ ( - Δ n ) . By Green’s Theorem Z ∂U f ( z ) dz = 0 . 2.6. The Residue Theorem and Its Applications 25 z 1 z 2 z N Γ U D Figure 2.2. The proof of Residue Theorem Hence Z γ f ( z ) dz = n X j =1 Z Δ j f ( z ) dz = n X j =1 (2 πi ) Res( f ; z 0 j ) . Applications of The Residue Theorem for Evaluating Integrals. We will apply the Residue Theorem to compute real variable integrals of the following type.  #### You've reached the end of your free preview.

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