Originally Posted by

**Beard** Hi, I have a cubic equation:

$\displaystyle z^3 + az^2 + bz + c = 0$

where a,b and c are real numbers. I have to show that if z is a solution then so is its complex conjugate, z*.

This much is true for ANY real polynomial $\displaystyle p(x)$ : its complex non-real roots come as conjugate pairs, because $\displaystyle p(\overline{z})=\overline{p(z)}$ , by the basic properties of the complex conjugation.

It is given that z = -2 -i is a solution to the equation. I must then write down a second complex solution.

If z = 2 is the thrid solution, what are a,b and c?

Then you know that the roots are $\displaystyle 2\,,\,-2-i\,,\,-2+i$ , and the either you compare coefficients of corresponding powers of x in the equality $\displaystyle ax^2+bx^2+cx+d=a(x-2)(x-(-2-i))(x-(-2+i))$ , or what is the same but more direct and simpler: you remember Viete's formulae

$\displaystyle a_1a_2a_3=-\frac{d}{a}\,,\,\,a_1a_2+a_1a_3+a_2a_3=\frac{c}{a} \,,\,\,a_1+a_2+a_3=-\frac{b}{a}$ , with $\displaystyle a_1,\,a_2,\,a_3$ the polynomial's roots.

Tonio

So far I have said that if z = x + iy, then the expansion of the first equation becomes

$\displaystyle (x^3 + 3x^2iy - 3xy^2 - iy^3) + a(x^2 + 2xiy - y^2) + b(x + iy) + c = 0$

when I do the same for the complex conjugate it just gets the x cubed term to become

$\displaystyle (x^3 + 3x^2iy + 3xy^2 + iy^3)$

but that isn't right because the both equations equate to zero and therefore the other, but they are not the same.

What do I do?

Any help appreciated.

(sorry moderators if this is in the wrong topic area)