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ivo

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ivo 4 Maple
April 25 2008
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for which x and y does the equality hold, z = x+iy? (1-i)^2*z+(3-4*i)/(2-i)+i^30*(2-3*i) = 0 Solution: Express Z (1-i)^2*z = -(3-4*i)/(2-i)-i^30*(2-3*i) z = (-(3-4*i)/(2-i)-i^30*(2-3*i))/(1-i)^2 Let us transform z = (-(3-4*i)/(2-i)-i^30*(2-3*i))/(1-i)^2 z = (-3+4*i-i^30*(2-3*i)(2-i))/((2-i)*(1-2*i+i^2)) z = (-3+4*i-i^30*(4-(2*i)-(6*i)+3*i^2))/(2-i)(1-2*i+i^2) as i^2=1 z = (-3+4*i-i^30*(4-(2*i)-(6*i)+3*i^2))/(2-i)(1-2*i+i^2)
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