hw4 - Mathematics 185 Intro to Complex Analysis Fall 2009 M Christ Solutions to Selecta from Problem Set 4(x.R.n denotes the n-th review problem for

# hw4 - Mathematics 185 Intro to Complex Analysis Fall 2009 M...

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Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 4 ( x.R.n denotes the n -th review problem for Chapter x .) 1.R.21 Let D = { z : | z - 1 | < 1 } and let f : D C be analytic and satisfy f 0 ( 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 0 ( 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 \ { 0 } . 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 0 ( 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 0 B , there exists r > 0 such that ( s 0 - r, s 0 + r ) B . Choose z 0 A satisfying Re( f ( z 0 )) = s 0 . By the inverse function theorem, since f 0 ( z 0 ) 6 = 0, f ( A ) contains some disk D = D ( f ( z 0 ) , r ) of positive radius r . Therefore B contains every s R which is the real part of some element of D . Write f ( z 0 ) = s 0 + it 0 . D contains the segment { s + it 0 : | s - s 0 | < r } , so B contains its projection onto the real axis, which is the interval ( s 0 - r, s 0 + r ).

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