solutionshw13-201141

solutionshw13-201141 - h can be extended to a holomorphic...

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Solutions homework 13. (1) Let G be open and connected and f,g analytic on G such that f ( z ) g ( z ) = 0 for all z G . Prove that either f ( z ) = 0 for all z G or g ( z ) = 0 for all z G . Solution: Assume neither f ( z ) = 0 for all z G or g ( z ) = 0 for all z G . Then f and g have each countably many zeros on G and thus also fg has countably many zeros, i.e. G is countable. Contradiction. (2) (Quals ’02) Let f,g : { z : | z | < 1 } → C be analytic functions such that | f ( z ) | = | g ( z ) | for all | z | < 1. Prove that every zero of g is also a zero of f of the same multiplicity and that thus f = λg for some λ with modulus one. Solution: Let z 0 be a zero of order m of g . Then g ( z ) = ( z - z 0 ) m g 1 ( z ), where g 1 ( z 0 ) 6 = 0. Hence f ( z ) ( z - z 0 ) m is bounded on B ( z 0 ,r ) for some r > 0. Hence z = z 0 is a removable singularity for f ( z ) ( z - z 0 ) m and lim z z 0 f ( z ) ( z - z 0 ) m 6 = 0. Therefore f ( z ) = ( z - z 0 ) m f 1 ( z ) with f 1 ( z 0 ) 6 = 0, i.e. f has a zero of multiplicity m at z = z 0 . Assume now that g is not identically zero. Then all the zeros of g are isolated and by the above h = f g has a removable singularity at each zero of g , i.e.
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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 ) = 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 ( f-λg )( z ) = 0 on B ( a,r ). As f-λg is holomorphic it follows that ( f-λg )( z ) = 0 on B (0 , 1). 1...
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