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Unformatted text preview: h can be extended to a holomorphic function on B (0 , 1) such that  h ( z )  = 1 for all z D (0; 1). Hence h ( z ) = with = 1. This implies that f ( z ) = g ( z ) for all z B (0 , 1). Second Solution for the second part: Assume g is not identically zero (otherwise f is identically zero too and any works). Then there exists a and r > 0 such that B ( a,r ) B (0 , 1) and g ( z ) 6 = 0 on B ( a,r ). Now h = f g is holomorphic on B ( a,r ) and satises  h ( z )  = 1 on B ( a,r ). Hence h is constant on B ( a,r ), so there exists with   = 1 such that h ( z ) = on B ( a,r ). Hence ( fg )( z ) = 0 on B ( a,r ). As fg is holomorphic it follows that ( fg )( z ) = 0 on B (0 , 1). 1...
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 Fall '11
 Schep

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