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but the integral is zero since <math> \nabla u </math> is zero. Hence <math> u(A) = u(B) </math> for any arbitrary point <math>A</math> and <math>B</math> and <math>u</math> is constant in <math>\Omega</math>.
 
but the integral is zero since <math> \nabla u </math> is zero. Hence <math> u(A) = u(B) </math> for any arbitrary point <math>A</math> and <math>B</math> and <math>u</math> is constant in <math>\Omega</math>.
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Practice Problems: Suggested Solutions/Thoughts
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[http://www.math.purdue.edu/~bell/MA425/prac1a.pdf| Link to the practice Problems]
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Problem 1.
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'''Daniel''': I interpret "an analytic function f has constant modulus on a domain" as a function f that maps all domain to some f(z) where |f(z)| = r. Then f is probably not constant, since this is a circle. So I am interpreting the problem wrong?
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Problem 2.
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'''Daniel''': Both a and b are simple - I would simply invoke the Fund. Theorem of Calc. Problem 2 could have been tricky if it involved going over the branch cut, but since it did not, I think we are safe. (check out pg. 175)
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Problem 3.
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'''Daniel''': I saw a similar problem on pg. 164. Shouldn't be too tricky since t is a real variable. Remembering what actually is <math>e^{3i\pi} </math> may be the hard part.
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Problem 4.
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'''Daniel''': Will return to it.
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Problem 5.
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'''Daniel''': hmm

Revision as of 17:04, 1 October 2012

Practice material for Exam 1 collaboration space

You can easily talk about math here, like this:

$ e^{i\theta} = \cos \theta + i \sin \theta. $

Is this the Cauchy Integral Formula?

$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a} \ dz $


This isn't directly related to the practice exam, but is concerning a fact discussed in class. In one of the first lessons an important fact was provided. Namely, Suppose u is continuously a differentiable function on a connected open set $ \Omega $ and that $ \nabla u \equiv 0 $ Then u must be constant on $ \omega $.

How/Why is

$ 0 = \int\nabla u\ ds $

If we have a domain, $ \Omega $ and $ \gamma $ a curve in $ \Omega $, where $ A, B $ are end points of $ \gamma $, from vector calculus, we have

$ u(B) - u(A) = \int_\gamma \nabla u \cdot ds $.

but the integral is zero since $ \nabla u $ is zero. Hence $ u(A) = u(B) $ for any arbitrary point $ A $ and $ B $ and $ u $ is constant in $ \Omega $.


Practice Problems: Suggested Solutions/Thoughts

Link to the practice Problems

Problem 1.

Daniel: I interpret "an analytic function f has constant modulus on a domain" as a function f that maps all domain to some f(z) where |f(z)| = r. Then f is probably not constant, since this is a circle. So I am interpreting the problem wrong?


Problem 2.

Daniel: Both a and b are simple - I would simply invoke the Fund. Theorem of Calc. Problem 2 could have been tricky if it involved going over the branch cut, but since it did not, I think we are safe. (check out pg. 175)


Problem 3. Daniel: I saw a similar problem on pg. 164. Shouldn't be too tricky since t is a real variable. Remembering what actually is $ e^{3i\pi} $ may be the hard part.


Problem 4. Daniel: Will return to it.


Problem 5. Daniel: hmm

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood