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Question: intermediate results

Question:intermediate results

ivo 4 Maple
example
April 25 2008
0
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) z = (-3+4*i+4-8*i-3)/(2-i)((1-(2*i)-1)) z = (-2-4*i)/(2-i)(-2*i) z = (-2-4*i)/(2-i)(-2*i) z = (-2-4*i)/(-2-4*i) z=1 x+iy = z Re(x+iy) = Re(1) x=1 Im(x+iy) = Im(1) y=0 as apparent from the (above) example, this is a simple algorithm: express via z separate the real and imaginary parts find the unknowns. The 50 problems in the problem set are all alike. How do you write a procedure in maple, which gives both the solution and the intermediate results?
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