Unformatted text preview: that ∑ n ≥  α nβ n  < ∞ . Show that Q n ≥ zα n zβ n converges normally on C \ { β n } (i.e. the complement of the closure of the set of β 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 ˆ f : U ∪ D ( p,r ) → C holomorphic, with ˆ f ± ± ± U = f . 1...
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 Spring '09
 Equals sign, converges, an  converges, Hadamard product formula, z−αn converges

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