f U V is said to be conformal or biholomoprhic if fis holomorphic one to one

# F u v is said to be conformal or biholomoprhic if fis

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f : U V is said to be conformal or biholomoprhic if f is holomorphic, one- to-one and onto. Examples of conformal mappings : 1. f : C C is conformal if and only if f is linear. 2. f : D (0 , 1) D (0 , 1) is conformal if and only if f is a Mobius transformation (a consequence of Schwarz-Pick Lemma). 3. f : { z | Imz > 0 } → D (0 , 1) is conformal if and only if f ( z ) = r z - a z - ¯ a , Ima > 0 . 4. f : { z | Imz > 0 } → { z | Imz > 0 } is conformal if and only if f ( z ) = az + b cz + d , where a, b, c, d are real numbers, and ad - bc > 0. 17 Normal Families : Let F = { f α } α Λ be a family of holomorphic functions on U C . We say that F is normal family if every sequence { f n } ⊂ F has a sequence that converges uniformly on every compact subsets of U . Montel’s Theorem Let F = { f α } α Λ be a family of holomorphic functions on U C that is bounded on compact subsets, i.e. for every compact set K U , there is a M K > 0 such that | f ( z ) | ≤ M k , z K, f ∈ F . Then F is normal. Riemann Mapping Theorem (Analytic Version) If U is a holomorphically simply connected open set in C and U 6 = C , then U is conformally equivalent to the unit disc . Idea of pf : Set, for a fixed P U , 1. F = { f | f : U D (0 , 1) is holomorphic, and one-to-one , f ( P ) = 0 } . 2. Use the normality to show that there is f 0 ∈ F such that | f 0 0 ( P ) | = sup f ∈F | f 0 ( P ) | . 3. Show that f 0 maps U onto the unit disc D (0 , 1). If U is simply connected, then U is a holomorphically simply connected. So the Riemann Mapping Theorem : If U is simply connected and U 6 = C . Then U is conformally equivalent to the unit disc . So the beauty of this theorem is that it basically says that the topological property of U implies the analytic property of U . Little Picard Theorem (Generalization of Liouville’s theorem) If f is entire and omits 2 values, then it is constant . Idea of pf : Step 1 . From omitting 2 values to omitting a lot of values (from spotting 2 cockroaches to a lot of cockroaches), this can be achieved by proving the following lemma: If U is simply connected and f is holomorphic on U and does not assume the values of 0 and 1 . Then there is a holomorphic curve on U such that f ( z ) = - exp( cosh[2 g ( z )]) . 18 Furthermore, it can be verified that g does not assume, on U , the values of π log( n + n - 1)+ 1 2 mπi for positive integer n and all integer m . Since these points from a vertices of a grid of rectangles with diagonal < 2, we have that g ( U ) contains no disc of radius 1 . Step 2 . We prove the so-called Bloch’s theorem Let f be analytic on a region containing D (0 , 1) and f (0) = 0 , f 0 (0) = 1 . Then there f ( D (0 , 1)) constains a disk of radius 1 / 72 (which is independent of f !) . Step 3 . The proof of little picard theorem can thus be proved by considering g ( z + z 0 ) /g 0 ( z 0 ) (by choosing some z 0 with g 0 ( z 0 ) 6 = 0 and by Bloch’s theorem, g ( D ( z 0 , R )) contains a disc of redius (1 / 72) R , which contains a disc of radius 1 when R is bigger enough, which contradicts with that fact that g ( U ) contains no disc of radius 1 derived in step 1. The proof of Bloch’s theorem : We first prove the following lemma: Lemma Let f be holomorphic on D (0 , 1) with f (0) = 0 , f 0 (0) = 1 . Assume that | f ( z ) | ≤ M for all z D (0 , 1) . Then M 1 and f ( D (0 , 1)) D (0 , 1 / (6 M )) .  #### You've reached the end of your free preview.

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