finals - Key to the Complex Analysis Final Exam, thanks to...

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Key to the Complex Analysis Final Exam, thanks to Da Zhang December 7, 2011 Part A: (1). ( Residue Formula ) Suppose f is a holomorphic function on a holomorphically simply connected domain U C , except for singularities z 1 , z 2 , ··· , z n , and γ is a piecewise closed C 1 curve in U , but not passing through any z k . Then the following equality holds: I γ f ( ζ ) = 2 πi n X k =1 Res f ( z k ) · n ( γ,z k ) where n ( k ) denotes the winding number of γ about each z k . Proof. U C is said to be holomorphically simply connected if every holomor- phic function on U has an anti-derivative (or primitive). Hence, H γ fdz = 0 if f is holomorphic on U (because let F with F 0 = f , then H γ = F ( γ ( b )) - F ( γ ( a )) = 0 since γ is closed. We’ll use this fact to prove the residue theroem. First, for each z k ,we set the principal part of the Laurent expansion of it to be S k , then consider the following auxiliary function: F ( z ) = f - ( S 1 + + S n ) , G ( z ) = S 1 + + S n Notice that G = ( S 1 + + S n ) is defined on U \{ z 1 , ,z n } , which is the same as f . Moreover, for each singularity z k , if we consider the Laurent expansion around it, the f - S k has just the regular part left, hence, by the Riemann’s theorem, z k is removable singularity for F = f - ( S 1 + + S n ); therefore, we can regard F as a holomorphic function on U . Next, by the simple connectedness of U , and the Cauchy’s theorem, we have that H γ F ( z ) dz = 0. Now, consider H γ G ( z ) dz = n k =1 H γ S k dz , we know that each S k is the principal part of the Laurent expansion of f at z k . So S k ( z ) = l =1 a k - l ( z - z k ) - l , and by 1
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the fact that γ is compact and does not pass through z k , we know l =1 a k - l ( z - z k ) - l converges uniformly on γ , thus, using the fact that H γ ( z - z k ) - k = 0 for k 6 = 1, I γ S k ( z ) dz = X l =1 I γ a k - l ( z - z k ) - l dz = I γ a k - 1 ( z - z k ) - 1 dz = 2 πiRes f ( z k ) · n ( γ,z k ) Therefore, in sum, combine the above results, we conclude that I γ f ( z ) dz = I γ ( F ( z ) + G ( z )) dz = I γ G ( z ) dz = n X k =1 I γ S k dz = 2 πi n X k =1 Res f ( z k ) · n ( k ) . This proved our desired Residue Formula. 2 Let g = f ( ζ ) / ( ζ - z 0 ). Observe that Res g ( z 0 ) = f ( z 0 ) by the Laurent expansion of g = f ( ζ ) / ( ζ - z 0 ) at the point z 0 . Apply the above residue theorem to g to yield that, for any (piecewise smooth) closed simple curve, γ with z 0 int ( γ ), since n ( 0 ) = 1, f ( z 0 ) = 1 2 πi I γ f ( ζ ) ( ζ - z 0 ) dζ. Thus the above Cauchy’s integral formula is proved. 2 To prove the argument principle, we first suppose f is a meromorphic function on a open set U , ¯ D ( P,r ) U , and f has no poles or zeros on ∂D ( ), then we shall show that 1 2 πi I ∂D ( P,r ) f 0 ( z ) f ( z ) dz = p X j =1 n j - q X k =1 m k where n 1 , ··· ,n p are the multiplicities of zeros z 1 , ,z p of f in D ( ), and m 1 , ,m p are the poles w 1 , ,w q of f in D ( ). To see this, for the function h ( z ) = f 0 ( z ) /f ( z ), we know that the possible singularities (actually poles) for it are z 1 , p and w 1 , q . Next, for any 2
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zero of f , z k , with multiplicity n k , on a neighborhood of which we can write f as f ( z ) = ( z - z k ) n k g ( z ), where g is holomorphic and has no zeros on the neighborhood. Then f 0 ( z ) f ( z ) = n k ( z - z k ) n k - 1 g ( z ) + ( z - z k ) n k g 0 ( z ) ( z - z k ) n k g ( z ) So it is easily seen that Res g ( z k ) = n k .
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finals - Key to the Complex Analysis Final Exam, thanks to...

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