solutionshw14-201141

# solutionshw14-201141 - Solutions Homework 14(1(Schwarz's...

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Solutions Homework 14. (1) (Schwarz’s lemma) Let f be a holomorphic function on B (0 , 1) with | f ( z ) | ≤ 1 for all | z | < 1 and f (0) = 0. a. Deﬁne f 1 ( z ) = f ( z ) z for z 6 = 0 in B (0 , 1). Prove that z = 0 is a removable singularity of f 1 . Solution: Since f is holomorphic at z = 0, lim z 0 f ( z ) z = f 0 (0) exists, so z = 0 is a removable singularity of f 1 ( z ). b. Prove that | f 1 ( z ) | ≤ 1 r on B (0 ,r ) for all 0 < r < 1. (Hint: use the maximum modulus principle.) Solution: If 0 < r < 1, then | f 1 ( z ) | = 1 r | f ( z ) | ≤ 1 r on | z | = r . Hence by the maximum modulus principle | f 1 ( z ) | ≤ 1 r for all | z | ≤ r . c. Conclude that | f ( z ) | ≤ | z | for all z B (0 , 1). Moreover if equality holds for some z 0 6 = 0, then there exists c with | c | = 1 such that f ( z ) = cz for all z B (0 , 1). Solution: Let z B (0 , 1). Then for all | z | ≤ r < 1 we have by part b) | f ( z ) | ≤ 1 r | z | . letting r 1 we get | f ( z ) | ≤ | z

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## This note was uploaded on 02/05/2012 for the course MATH 703 taught by Professor Schep during the Fall '11 term at South Carolina.

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solutionshw14-201141 - Solutions Homework 14(1(Schwarz's...

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