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| + | '''Problem 8:''' My first map was <math>z^4</math> to get it to the unit circle. After that I remembered that since <math>exp(z)</math> took vertical lines to circles, <math>log(z)</math> should take circles back to those lines. But I recall from lectures that <math>exp(z)</math> gives 1 map, but <math>log(z)</math> gives infinitely many. Does that apply in this case? Is <math>log(z^4)</math> correct? --[[User:Aata|Aata]] | ||
Revision as of 09:17, 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
Problem 8: My first map was $ z^4 $ to get it to the unit circle. After that I remembered that since $ exp(z) $ took vertical lines to circles, $ log(z) $ should take circles back to those lines. But I recall from lectures that $ exp(z) $ gives 1 map, but $ log(z) $ gives infinitely many. Does that apply in this case? Is $ log(z^4) $ correct? --Aata
