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 ) = λ with λ = 1. This implies that f ( z ) = λg ( 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 λ 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 satisﬁes  h ( z )  = 1 on B ( a,r ). Hence h is constant on B ( a,r ), so there exists λ with  λ  = 1 such that h ( z ) = λ on B ( a,r ). Hence ( fλg )( z ) = 0 on B ( a,r ). As fλg is holomorphic it follows that ( fλg )( z ) = 0 on B (0 , 1). 1...
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 Fall '11
 Schep
 Taylor Series, Complex number, Analytic function, Holomorphic function

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