.Hi, I have a cubic equation:
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 : its complex non-real roots come as conjugate pairs, because , 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 , and the either you compare coefficients of corresponding powers of x in the equality , or what is the same but more direct and simpler: you remember Viete's formulae
, with the polynomial's roots.
So far I have said that if z = x + iy, then the expansion of the first equation becomes
when I do the same for the complex conjugate it just gets the x cubed term to become
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)