sol108c2

sol108c2 - MA108C HW 2 SOLUTIONS PAUL NELSON AND FOKKO VAN...

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MA108C HW 2 SOLUTIONS PAUL NELSON AND FOKKO VAN DE BULT (1) Applying the Cauchy integral formula to f ( z ) = z ( z + 3) / ( z - 8) at the point z = - i D (2 , 4) we get 1 2 πi I ∂D (2 , 4) f ( ζ ) ζ + i = f ( - i ) . Note that f ( z ) is holomorphic on D (2 , 4) (its only singularity is outside this disc). (2) Put f ( z ) = e - ( a - ib ) z , so that e - ax cos( bx ) = < f ( x ). Put c = a 2 + b 2 and choose ω ( - π/ 2 ,π/ 2) so that e ( a - ib ) = c , i.e. e = a + ib a 2 + b 2 . Fix R > 0, and let γ 1 ( t ) = t for t [0 ,R ], γ 2 ( t ) = Re it for t [0 ], and γ 3 ( t ) = e t for t [0 ,R ], regarding each γ i as a contour. Then ( R γ 1 + R γ 2 - R γ 3 ) f ( z ) dz = 0 by Cauchy’s theorem. As R → ∞ , we have < ( γ 2 ( t )) R cos a for t [0 ], so | R γ 2 f ( z ) dz | = O ( Re - R cos a ) = o (1), and R γ 3 f ( z ) dz = e R R 0 e - ct dt = e c + o (1), whence R R 0 e - ax cos( bx ) dx = < ( R γ 1 f ( z ) dz ) = cos
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This document was uploaded on 12/08/2010.

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