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Unformatted text preview: ) has a zero or pole a at the boundary of D we delete from D its intersection with a small circle of radius r with center at a . Fig.1 Applying the Cauchy Residue Theorem we obtain 1 2 i Z P f d f = X P ( f ) = X P ( f ) m . When we integrate over the part P 1 of the boundary dened by Im = h we obtain 1 2 i Z P 1 f dz f = ( f ) . In fact, considering the Fourier expansions of f at , we get f ( ) = X n = ( f ) a n e 2 in , f ( ) = X n = ( f ) (2 in ) a n e 2 ni ....
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.
 Fall '09
 ONTONKONG
 Algebra

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