Isolated Singulari Anf taie con tinction 展开定理( Laurent) (要点) f()=∑a1(-B2<|-M<R2 2πiJe(-b)n dc 对于沿C1的积分 2-20=2i,(2-b)-(-b)
Expansion in Laurent Series Isolated Singularities of Uniform Function Analytic Continuation Theorem (Laurent) Illustrative Examples Laurent Expansion: Multivalued Functions Ðm½n(Laurent) (:) f(z)= X∞ n=−∞ an(z−b) n R1 <|z−b|<R2 an = 1 2π i I C f(ζ) (ζ−b) n+1 dζ éu÷C1È© − 1 2π i I C1 f(ζ) ζ−z dζ = 1 2π i I C1 f(ζ) (z−b) − (ζ−b) dζ C. S. Wu 1lù )Û¼êLaurentÐm
Isolated Singulari Anf taie con tinction 展开定理( Laurent) (要点) f()=∑a1(-B2<|-M<R2 2πiJe(-b)n dc 对于沿C1的积分 f()1f()、/-b 2πic1-z 2π1JC1 bz(2-b d
Expansion in Laurent Series Isolated Singularities of Uniform Function Analytic Continuation Theorem (Laurent) Illustrative Examples Laurent Expansion: Multivalued Functions Ðm½n(Laurent) (:) f(z)= X∞ n=−∞ an(z−b) n R1 <|z−b|<R2 an = 1 2π i I C f(ζ) (ζ−b) n+1 dζ éu÷C1È© − 1 2π i I C1 f(ζ) ζ−z dζ = 1 2π i I C1 f(ζ) z−b X ∞ k=0 ζ−b z−b k dζ C. S. Wu 1lù )Û¼êLaurentÐm
Isolated Singulari Anf taie con tinction 展开定理( Laurent) (要点) f()=∑a1(-B2<|-M<R2 2πiJe(-b)n dc 对于沿C1的积分 f() S-b 2丌 =1f() 2TiJG, 2-b b d ∑|np(-yd(c--k-1 k=0 (|z-b>R1) 第八讲解析函数的 Laurent
Expansion in Laurent Series Isolated Singularities of Uniform Function Analytic Continuation Theorem (Laurent) Illustrative Examples Laurent Expansion: Multivalued Functions Ðm½n(Laurent) (:) f(z)= X∞ n=−∞ an(z−b) n R1 <|z−b|<R2 an = 1 2π i I C f(ζ) (ζ−b) n+1 dζ éu÷C1È© − 1 2π i I C1 f(ζ) ζ−z dζ = 1 2π i I C1 f(ζ) z−b X ∞ k=0 ζ−b z−b k dζ = X ∞ k=0 1 2π i I C1 f(ζ)(ζ−b) k dζ (z−b) −k−1 (|z − b| > R1) C. S. Wu 1lù )Û¼êLaurentÐm
Isolated Singulari Anf taie con tinction 展开定理( Laurent) (要点) f()=∑a1(-B2<|-M<R2 2πiJe(-b)n dc 对于沿C1的积分 f() S-b 2πic1-z dc- 2Ti JCi do b k=0 ∑ f() b) (|z-b>R1) 第八讲解析函数的 Laurent展开
Expansion in Laurent Series Isolated Singularities of Uniform Function Analytic Continuation Theorem (Laurent) Illustrative Examples Laurent Expansion: Multivalued Functions Ðm½n(Laurent) (:) f(z)= X∞ n=−∞ an(z−b) n R1 <|z−b|<R2 an = 1 2π i I C f(ζ) (ζ−b) n+1 dζ éu÷C1È© − 1 2π i I C1 f(ζ) ζ−z dζ = 1 2π i I C1 f(ζ) z−b X ∞ k=0 ζ−b z−b k dζ = X −∞ n=−1 1 2π i I C1 f(ζ) (ζ−b) n+1dζ (z−b) −n (|z − b| > R1) C. S. Wu 1lù )Û¼êLaurentÐm
Isolated Singulari Anf taie con tinction 展开定理( Laurent) (要点) f(2)=∑an(2-b)R1<|2-b<R2 2πi(-b)n+1 dc 对于沿C2的积分,可直接引用 Taylor展开的结果 1f(G) 1(f( S-b do ∑ f(s 2丌 (S-6)n+ids/(2-b) (|z-b<R2)
Expansion in Laurent Series Isolated Singularities of Uniform Function Analytic Continuation Theorem (Laurent) Illustrative Examples Laurent Expansion: Multivalued Functions Ðm½n(Laurent) (:) f(z)= X∞ n=−∞ an(z−b) n R1 <|z−b|<R2 an = 1 2π i I C f(ζ) (ζ−b) n+1 dζ éu÷C2È©§Ú^TaylorÐm(J 1 2π i I C2 f(ζ) ζ−z dζ = 1 2π i I C2 f(ζ) ζ−b X ∞ n=0 z−b ζ−b n dζ = X ∞ n=0 1 2π i I C2 f(ζ) (ζ−b) n+1dζ (z−b) n (|z − b| < R2) C. S. Wu 1lù )Û¼êLaurentÐm