This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 4 ( x.R.n denotes the nth review problem for Chapter x .) 1.R.21 Let D = { z :  z 1  < 1 } and let f : D → C be analytic and satisfy f ( z ) = 1 /z and f (1) = 0. Show that f ( z ) = log( z ). Solution. Let L ( z ) denote the unique logarithm of z with argument in ( π/ 2 ,π/ 2), for z ∈ C \ (∞ , 0]. I will show more precisely, and less ambiguously, that f ( z ) = L ( z ) for all z ∈ D . Consider g ( z ) = f ( z ) L ( z ) for z ∈ D . Then g ( z ) ≡ 0. Since D is connected, g is constant. Therefore f ( z ) L ( z ) ≡ f (1) L (1) = 0 0 = 0. 1.R.26. Let g : A → C be an analytic function on any open set A . Let B = { z ∈ A : g ( z ) 6 = 0 } . Show that B is open, and that 1 /g is an analytic function on B . Solution. Since g is analytic, it is continuous. Given any z ∈ B , define ε =  g ( z )  / 2. There exists δ > 0 such that  w z  < δ ⇒  g ( w ) g ( z )  < ε . By the triangle inequality, for any such w ,  g ( w )  ≥  g ( z )    g ( w ) g ( z )  ≥ 1 2  g ( z )  . Therefore B contains the open disk D ( z,δ ). By definition of an open set, B is open. Define G ≡ g , but with domain restricted to B . Then G is analytic. Define h ( w ) = 1 /w for all w ∈ C \ { } . Then h is analytic, and its domain contains the range of G . Since 1 /g = h ◦ G is the composition of two analytic functions, it is also analytic. 1.R.28 Let f : A → C be analytic and assume that f ( z ) 6 = 0 for every z ∈ A . Prove that { Re( f ( z )) : z ∈ A } is an open subset of R . Solution. Let B = { Re( f ( z )) : z ∈ A } . In order to prove that B is an open subset of R , we must show that for any s ∈ B , there exists r > 0 such that ( s r,s + r ) ⊂ B . Choose z ∈ A satisfying Re( f ( z )) = s . By the inverse function theorem, since f ( z ) 6 = 0, f ( A ) contains some disk D = D ( f ( z ) ,r ) of positive radius r . Therefore B contains every s ∈ R which is the real part of some element of D . Write f ( z...
View
Full
Document
This note was uploaded on 10/10/2009 for the course MATH 185 taught by Professor Lim during the Spring '07 term at Berkeley.
 Spring '07
 Lim
 Math

Click to edit the document details