hwsolB01 - Therefore v = 1*(cos(pi) + j*sin(pi)) = -1 + j*0...

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Sheet1 Page 1 HW 1 SOLUTION _____________ (i) Plot (-1,sqrt(3)) and (1,-1) on the (x,y) plane. (ii) |z1| = sqrt((-1)^2+(sqrt(3))^2) = sqrt(1+3) = 2 angle(z1) = arctan(-sqrt(3)) + pi = 2*pi/3 = 2.0944 |z2| = sqrt(1+1) = sqrt(2) = 2.828 angle(z2) = arctan(-1/1) = -pi/4 (iii) z1^3 = (-1+sqrt(3)*j)*(-1+sqrt(3)*j)*(-1+sqrt(3)*j) = (-1+sqrt(3)*j)*(1-sqrt(3)*j-sqrt(3)*j+3*j^2) = (-1+sqrt(3)*j)*(-2-2*sqrt(3)*j) = 2-2*sqrt(3)*j+2*sqrt(3)*j-6*j^2 = 8 z2^2 = (1-j)*(1-j) = 1-j-j+j^2 = -2j z1^2 = (-1+sqrt(3)*j)*(-1+sqrt(3)*j) = 1-sqrt(3)*j-sqrt(3)*j+3*j^2 = -2-2*sqrt(3)*j z1^2 + 2*z1 + 4 = -2-2*sqrt(3)*j -2 +2*sqrt(3)*j + 4 = 0 (iv) v = (z1)^6*(z2)^(-12) |v| = |z1|^6*|z2|^(-12) = 2^6 * (sqrt(2))^(-12) = 1 angle(v) = 6*angle(z1) - 12*angle(z2) = 6*(2*pi/3) - 12*(-pi/4) = 4*pi+3*pi = 7*pi (same as pi = 3.14159265)
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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.

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hwsolB01 - Therefore v = 1*(cos(pi) + j*sin(pi)) = -1 + j*0...

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