Question

((z-2/6z+(z-2)^2)+((z+4)^2-12/z^3-2^3)-(1/z-2)):z^3+2z^2+2z+4/z^3-2z^2+2z-4

Answer 1

2z^4 - 3z^3 + 4z^2 - 3z + 2 = 0 
=> Dividing by z^2, 
2z^2 - 3z + 4 - 3/z + 2/z^2 = 0 
=> 2 (z^2 + 1/z^2) - 3(z + 1/z) + 4 = 0 
=> 2 [(z + 1/z)^2 - 2] - 3(z + 1/z) + 4 = 0 
=> 2(z + 1/z)^2 - 3(z + 1/z) = 0 
=> (z + 1/z) [2(z + 1/z) - 3] = 0 
=> (z^2 + 1) (2z^2 - 3z + 2) = 0 
=> (z^2 + 1) (2z^2 - 3z + 2) = 0 
=> z^2 + 1 = 0 or 2z^2 - 3z + 2 = 0 
=> z = ± i or z = (1/2) [3 ± √(9 - 16)] 
=> z = ± i or z = (3/2) ± i (7/2)