mid1 - MATH 6321 First Midterm February 18, 2009 You can...

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Unformatted text preview: MATH 6321 First Midterm February 18, 2009 You can use your book and notes. No laptop or wireless devices allowed. Write clearly and try to make your arguments as linear and simple as possible. The complete solution of one exercise will be considered more that two half solutions. All numbers appearing in the test are complex numbers and all functions are from C to C . You should solve at least one among ex. 1 and 2, one among 3 and 4, and one among 5 and 6. A total of 50 pts will give you an A. Name: Question: 1 2 3 4 5 6 7 Total Points: 10 10 10 10 10 10 20 80 Score: MATH 6321 First Midterm February 18, 2009 1. (10 points) Find a one-to-one analytic function f mapping the open unit disk D = { z || z | < 1 } (1) to the disk C = { z || z- 1- 2 i | < 2 } (2) such that f (0) = 1 + i and f (0) is real and positive. ( Hint : you may use the result of exercise III.3.9) Solution: We first note that the function g ( z ) = 2 z + 1 + 2 i maps D into C . If h ( z ) is a Mobius transformation that maps D into itself, f ( z ) = g h ( z ) maps D into C . From III.3.9 we know that h ( z ) = az + b bz + a (3) We thus have that f ( z ) = 2 az + 2 b bz + a + 1 + 2 i (4) We get f (0) = 2 b a + 1 + 2 i 2 b a =- i f (0) = 2 | a | 2- 2 | b | 2 a 2 | a | 2- | b | 2 a 2 > (5) The second line implies that a is either real or pure immaginary. We can thus reduce the choice to a = 1 or a = i . But the first line tells that 2 | b | = | a | thus a = 1 and b =- i/ 2. Page 1 of 7 MATH 6321 First Midterm February 18, 2009 2. (10 points) Construct a Mobius transformation that maps the circle | z | = 1 and | z- 1 4 | = 1 4 onto two concentric circles. ( Hint : find two points z and z * that are conjugate with respect to both circles and ...) Solution: By symmetry we have that if z and z * are conjugate with respect to both circles they have to be real. This implies that they satisfy the equation: z- 1 4 1 z- 1...
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mid1 - MATH 6321 First Midterm February 18, 2009 You can...

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