solutionshw14-201141

solutionshw14-201141 - Solutions Homework 14. (1)...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

solutionshw14-201141 - Solutions Homework 14. (1)...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online