814F07xm1sol

# 814F07xm1sol - Math 814 Exam 1 Due Nov 7 2007 You may...

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Math 814 Exam 1 Due: Nov 7, 2007 You may consult books, homework solutions or the instructor, but no one else. All work must be your own. 1. Let f ( z ) = z 2 +1 . Find equations for and sketch the curves obtained as images of horizontal and vertical lines under f ( z ) . We have f = u + iv where u = x 2 - y 2 +1 and v = 2 xy . First, f ( R ) = [1 , ) and f ( i R ) = ( -∞ , 1] . If x = a 6 = 0 is constant we have u = 1 + a 2 - y 2 , v = 2 ay, so u = 1 + a 2 - v 2 4 a 2 . In the ( u, v ) plane, this is a left-opening parabola with vertex (1 + a 2 , 0) . As | a | increases, the vertex moves to the right and the parabola opens wider, becoming more like a straight line. If y = b 6 = 0 is constant we get u = 1 - b 2 + v 2 4 b 2 . In the ( u, v ) plane, this is a right-opening parabola with vertex (1 - b 2 , 0) . As | b | increases, the vertex moves to the left and the parabola opens wider, becoming more like a straight line. Each of these parabolas is perpendicular to all of the previous ones. 2. Repeat Problem 1 for f ( z ) = cos z . We have

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## This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

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814F07xm1sol - Math 814 Exam 1 Due Nov 7 2007 You may...

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