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Unformatted text preview: Therefore v = 1*(cos(pi) + j*sin(pi)) = -1 + j*0 (v) |z-z1-z2|=|z1| is a circle of radius |z1| = sqrt(1+3) = 2 with center at the point z1+z2 = (0,sqrt(3)-1). Note that z=z2 lies on that circle. If z is the point of intersection of the circle and the real axis, then z = x + j*0 = x (i.e., z lies on the real axis) Sheet1 Page 2 and |x + (sqrt(3)-1) *j| = 2 Squaring both sides, x^2 + (sqrt(3)-1)^2 = 4 which has roots x = -1.8612 and x = 1.8612 x If z is the point of intersection of the circle and the imaginary axis, then z = j*y (i.e., z lies on the imaginary axis) and |y*j + (sqrt(3)-1) *j| = 2 Squaring both sides, (y + sqrt(3)-1)^2 = 4 which has roots y = 2+1-sqrt(3) = 3-sqrt(3) and y = 2-1+sqrt(3) = 1+sqrt(3)...
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This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.
- Spring '08