This preview shows page 1. Sign up to view the full content.
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  zz  < 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 1z = X n =0 z n we have 1 z 21 =1 1z 2 = X n =0 z 2 n . Its radius of convergence is R = 1 , since the function 1 / ( z 21) is analytic in the disc  z  < 1 , but not at z = 1 . 1...
View Full
Document
 Fall '07
 Lim
 Power Series

Click to edit the document details