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 =\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 
, and the either you compare coefficients of corresponding powers of x in the equality (x-(-2-i))(x-(-2+i)))
, or what is the same but more direct and simpler: you remember Viete's formulae 
, with 
the polynomial's roots. Tonio
So far I have said that if z = x + iy, then the expansion of the first equation becomes
 + 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
)
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)
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Originally Posted by Beard 
