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 3 1.5.15 Let f be an analytic function on D = { z :  z  < 1 } . Suppose that Re( f ( z )) = 3 for all z D . Prove that f is constant. Solution. Write f = u + iv where u,v are realvalued. The hypothesis is that u 3 on D . Therefore the partial derivatives satisfy u x u y 0 on D . By the CauchyRiemann equations, v y = v x = 0 at every point of D . Since f = u x + iv x , f ( z ) = 0 for all z D . Since D is connected, this forces f to be constant. Comment. D is convex, and therefore is pathconnected; if z,w D then the line segment joining z,w belongs to D . Any pathconnected set is connected. 1.5.19(c) Let f ( x ) = z +1 z 1 . What is the image of the y axis under f ? Solution. Let y R . Let = { z C :  z  = 1 } . f ( iy ) = 1+ iy 1 iy . Note that if w is the numerator 1 + iy , then the denominator 1 iy equals w . Since  w  =  w  for any complex number, both numerator and denominator have the same absolute values. Thus the ratio has absolute value 1, hence is an element of . Thus f ( i R ) . We calculate f ( iy ) = (1 + iy ) 2 (1 + iy )(1 iy ) = 1 y 2 1 + y 2 + i 2 y 1 + y 2 = h ( y ) + ig ( y ) . I claim that as y varies over [0 , 1], g ( y ) = 2 y/ (1 + y 2 ) varies over [0 , 1]; g : [0 , 1] [0 , 1] is a bijection. This is a simple one variable calculus exercise; I leave the details to you. Next, I claim that y 7 f ( iy ) is a bijection from [0 , 1] to the quarter of which lies in the closed first quadrant Q 1 = { x + iy : x 0 and y } . Note that h ( y ) 0 and g ( y ) for y [0 ,...
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 University of California, Berkeley.
 Spring '07
 Lim
 Math

Click to edit the document details