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Unformatted text preview: z = i , with residues cos i/ 2 i at z = i and cos(-i ) /-2 i at z =-i . Since these residues add up to zero (cosine is an even function) and both poles are inside the contour, the integral is zero. 4. Suppose that, for some z C and some > 0 , the open set | z-z | < is disjoint from the image of f . Let g ( z ) = 1 f ( z )-z . Then g is entire, and | g ( z ) | 1 / for all z , so by Liouvilles theorem g is constant. This implies that f is constant, too. Thus we have shown the contrapositive of the desired result. 5. Using the series 1 1-z = X n =0 z n we have 1 z 2-1 =-1 1-z 2 =- X n =0 z 2 n . Its radius of convergence is R = 1 , since the function 1 / ( z 2-1) is analytic in the disc | z | < 1 , but not at z = 1 . 1...
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- Fall '07
- Power Series