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'''Problem 1''': I am still a bit confused as to the geometric interpretation. If <math>|f(z)|=c</math> then doesn't this imply that f(z) could map the complex plane to a circle of radius <math>c</math>? Also if <math>f(z)=c</math> then this means that the complex plane maps to a single point, correct? I've gone over the symbolic manipulation using the C-R-Equations and everything works out fine, but it doesn't completely convince me. Is it because the function is analytic?--[[User:Rgilhamw|Rgilhamw]] 12:05, 6 October 2009 (UTC) | '''Problem 1''': I am still a bit confused as to the geometric interpretation. If <math>|f(z)|=c</math> then doesn't this imply that f(z) could map the complex plane to a circle of radius <math>c</math>? Also if <math>f(z)=c</math> then this means that the complex plane maps to a single point, correct? I've gone over the symbolic manipulation using the C-R-Equations and everything works out fine, but it doesn't completely convince me. Is it because the function is analytic?--[[User:Rgilhamw|Rgilhamw]] 12:05, 6 October 2009 (UTC) | ||
| + | Yes, I believe f(z) cannot map the complex plane to a circle because any bounded analytic function on the entire plane must be constant by [http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29 Loiuville's Theorem]. The linked proof is "algebraic" but it does confirm your suspicion that f(z) cannot map to a circle due to it's "analytic" restriction. | ||
| + | --[[User:Ysuo|Ysuo]] | ||
Professor Bell, | Professor Bell, | ||
Revision as of 11:01, 6 October 2009
Discussion area to prepare for Exam 1
Problem 1: I am still a bit confused as to the geometric interpretation. If $ |f(z)|=c $ then doesn't this imply that f(z) could map the complex plane to a circle of radius $ c $? Also if $ f(z)=c $ then this means that the complex plane maps to a single point, correct? I've gone over the symbolic manipulation using the C-R-Equations and everything works out fine, but it doesn't completely convince me. Is it because the function is analytic?--Rgilhamw 12:05, 6 October 2009 (UTC)
Yes, I believe f(z) cannot map the complex plane to a circle because any bounded analytic function on the entire plane must be constant by Loiuville's Theorem. The linked proof is "algebraic" but it does confirm your suspicion that f(z) cannot map to a circle due to it's "analytic" restriction.
--Ysuo
Professor Bell, For problem 7 on the practice problem worksheet, would it be valid to just let be z equal to the curve R*exp(it) in the integrand and take the limit as R goes to infinity, of the integral showing that the integrand approaches 0 and thus the integral goes to 0?--Adrian Delancy
Adrian, no that isn't enough because the length of the curve goes to infinity at the same time that the integrand goes to zero. It is a more subtle problem and you need to use the estimate that I used in class today. --Steve Bell See my notes at
