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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</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| | + | '''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|Robert Gilham-Westerman]] 12:05, 6 October 2009 (UTC) |
| − | + | Robert, another way to state this fact is to say that if f(z) is analytic on the complex plane and maps the plane INTO a circle of radius c, then f must be constant. You can prove this by means of the Cauchy-Riemann equations even if the plane is replaced by a disc (or more generally by a connected open set). --[[User:Bell|Steve Bell]] | |
| − | --[[User:Ysuo| | + | 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 Liouville'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|Yu Suo]] |
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| + | Yu, Liouville's Theorem is over kill here. The fact follows from the Cauchy-Riemann equations and it all works even if the whole plane is replaced by a disc. --[[User:Bell|Steve Bell]] | ||
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[http://www.math.purdue.edu/~bell/MA425/Lectures/lec10-05.pdf lec10-05.pdf] | [http://www.math.purdue.edu/~bell/MA425/Lectures/lec10-05.pdf lec10-05.pdf] | ||
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| + | Professor Bell, in class on Monday you went through a brief explanation of problem 2b on the practice problems and you chose the branch cut of log with arg = -pi/2, but I assume that any argument is ok as long as they fit the description? | ||
| + | --[[User:Ysuo|Yu Suo]] 16:55, 6 October 2009 (UTC) | ||
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| + | Yu, yes, you just need to pick a branch cut that stays out of the path of the curve. --[[User:Bell|Steve Bell]] | ||
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| + | Choosing the branch cut of <math>arg z=-\frac{\pi}{2}</math> determines the range of valid <math>\theta</math> for evaluating the log function. It is necessary to choose a branch cut with an arg between <math>0 + n2\pi</math> and <math>-\pi + n2\pi</math> because the curve is constrained to the upper half of the plane, and choosing a branch of log in this range ensures that the log function will remain analytic over the open set we are interested in (i.e. the upper half plane). --[[User:Rgilhamw|Robert Gilham-Westerman]] 20:18, 6 October 2009 (UTC) | ||
Latest revision as of 08:12, 7 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?--Robert Gilham-Westerman 12:05, 6 October 2009 (UTC)
Robert, another way to state this fact is to say that if f(z) is analytic on the complex plane and maps the plane INTO a circle of radius c, then f must be constant. You can prove this by means of the Cauchy-Riemann equations even if the plane is replaced by a disc (or more generally by a connected open set). --Steve Bell
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 Liouville'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. --Yu Suo
Yu, Liouville's Theorem is over kill here. The fact follows from the Cauchy-Riemann equations and it all works even if the whole plane is replaced by a disc. --Steve Bell
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
Professor Bell, in class on Monday you went through a brief explanation of problem 2b on the practice problems and you chose the branch cut of log with arg = -pi/2, but I assume that any argument is ok as long as they fit the description? --Yu Suo 16:55, 6 October 2009 (UTC)
Yu, yes, you just need to pick a branch cut that stays out of the path of the curve. --Steve Bell
Choosing the branch cut of $ arg z=-\frac{\pi}{2} $ determines the range of valid $ \theta $ for evaluating the log function. It is necessary to choose a branch cut with an arg between $ 0 + n2\pi $ and $ -\pi + n2\pi $ because the curve is constrained to the upper half of the plane, and choosing a branch of log in this range ensures that the log function will remain analytic over the open set we are interested in (i.e. the upper half plane). --Robert Gilham-Westerman 20:18, 6 October 2009 (UTC)
