Question

The cubic equation z^3+Az^2+Bz+26=0, A, BeR(A, B are elements in the set R) one root of the above cubic equation is 1+i. Find the real root (ii) Determine the values of A and B.

Answer 1

First of all, as coefficients A and B are real, we know that if $z \in \mathbb{C}$ is a root of an equation, its' complex conjugate $\bar{z}$ is also a root of this equation :

$\bar{z}^3+A\bar{z}^2+B\bar{z}+26=\bar{z^3}+\bar{(Az^2)}+\bar{(Bz)}+26=\bar{0}=0$

Therefore, $\bar{(1+i)} = 1-i$ is also a root of the equation.

Now, to find the third root we can apply, for example, Vieta's formulas :

$z_1\times z_2 \times z_3 = (-1)^3 26$

$(1-i)\times(1+i)\times z_3 = -26$

$z_3 = -13$

To find the coefficients we can either replace $z$ in the equation by the roots we've found (which is more direct method), either reapply the Vieta's formulas :

$A = -z_1-z_2-z_3=11$

$B = z_1z_2+z_2z_3+z_1z_3 = -24$

We can even replace $z$ in our equation by its' 3 different values and verify that these calculations are correct.