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Unformatted text preview: that n  n n  < . Show that Q n z n z n converges normally on C \ { n } (i.e. the complement of the closure of the set of n s). Now let U C be open, and take a sequence { c j } U , which is dense in U . Let { a j } be the sequence a j = c jb j1 c 2 , i.e. c 1 ,c 1 ,c 2 ,c 3 ,c 1 ,c 2 ,c 3 ,c 4 ,c 5 ,c 1 ,... and assume b j U are such that n  a nb n  < . Show that f ( z ) = Q n zb j za j is holomorphic on U and no p U is regular for f , i.e. for any p U there does not exist any r > 0 and f : U D ( p,r ) C holomorphic, with f U = f . 1...
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This document was uploaded on 12/08/2010.
 Spring '09

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