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...

This preview shows page 1 - 2 out of 4 pages.

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 ).
Image of page 1
Image of page 2

You've reached the end of your free preview.

Want to read all 4 pages?

  • Fall '07
  • Lim
  • Math, Topology, Continuous function, Metric space, Jordan Curve Theorem

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors