Answer to Question #141156 in Complex Analysis for Mac Evans

Question

Given the cubic equation 2z^3-z^2+4z+p=0, peR( p is an element of the set of real numbers)
z=1+2i (i) Find the other solution (ii) Determine the value of P

Answer 1

$\displaystyle f(z) = 2z^3-z^2+4z+p = 0\\ \textsf{By the factor theorem,}\, f(1 + 2i) = 0\\ 2(1 + 2i)³ - (1 + 2i)² + 4(1 + 2i) + p = 0\\ 2(1 + 6i - 12 - 8i) - (1 + 4i - 4) + 4 + 8i + p = 0\\ 2(-2i - 11) - (-3 + 4i) + 4 + 8i + p = 0\\ -4i - 22 + 3 - 4i + 4 + 8i + p = 0\\ \therefore p = 15\\ \textsf{By the Fundamental theorem of Algebra}\\ \textsf{the conjugate of}\, 1 + 2i \, \textsf{is also a root of}\, f(z)\\ \therefore (z - (1 + 2i))(z - (1 - 2i)) = 0\\ z^2 - (1 + 2i + 1 - 2i)z + (1 + 4) = 0\\ z^2 - 2z + 5 = 0\\ \textsf{By Long division, we obtain the last factor}\\ 2z + 3\\ z^2 - 2z + 5|2z^3-z^2+4z+15\\ -(2z^3 - 4z^2 + 10z)\\ 3z^2 - 6z + 15\\ -(3z^2 - 6z + 15) = 0\\ \implies(z^2 - 2z + 5)(2z + 3) = 0\\ 2z + 3 = 0 \implies z = -\frac{3}{2}\\ \therefore z = -\frac{3}{2}\, \textsf{is another solution.}\\ (i)\\ -\frac{3}{2}\\ (ii)\\ p = 15$