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[http://www.math.purdue.edu/~bell/MA425/prac1.pdf Practice Problems for Exam 1] | [http://www.math.purdue.edu/~bell/MA425/prac1.pdf Practice Problems for Exam 1] | ||
| − | '''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< | + | '''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) |
Revision as of 07:05, 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)
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
