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Unformatted text preview: The Little Picard Theorem and Its Proof November 22, 2011 Little Picard Theorem. If f is entire and its image omits 2 values, then it is constant . Lemma 1. Let U be a holomorphically simply connected domain. Let f be holomorphic on U and its image omits the value of and 1 . Then there exists a holomorphic g on U such that f ( z ) = exp( iπ cosh[2 g ( z )]) and g ( C ) contains no disc of radius 1. it Proof. The function g indeed can be constructed by taking (a branch) of “log” and “square root”. Here is the construction: Since f never vanishes and U is holomorphically simply connected, there is a holomorphic function h on U such that e h = f on U . Let F := 1 2 πi h . Since f omits the value 1, F does not assume any integer values, inparticular F omits the value 0 and 1. So there are holomorphic functions H 1 ,H 2 on U (again, use the fact that U is holomorphically simply connected) such that H 2 1 = F,H 2 2 = F 1. Let H = H 1 H 2 , then H is never zero on U , so there is a holomorphic function g such that e g = H . From the definition cosh(2 g ) + 1 = 1 2 ( e 2 g + e 2 g ) + 1 = 1 2 ( e g + e g ) 2 = 1 2 ( H + H 1 ) 2 = 2 F. Thus f ( z ) = exp( iπ cosh[2 g ( z )]) . Next, we claim that g does not assume, on U , the values of π log( √ n + √ n 1)+ 1 2 mπi for all positive integers n and all integers m due to the fact that the image of f omits the value 1. Indeed, if there isomits the value 1....
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This note was uploaded on 02/21/2012 for the course MATH 4377 taught by Professor Staff during the Summer '08 term at University of Houston.
 Summer '08
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