Roots of a cubic equation

Question

I have the following equation:

$s^3+as+b=0$

Now I want the values for a and b for which the given equation has the following complex roots:

$c \pm di$

I don't really care about the remaining root.

Any ideas? I have Matlab available, but I don't know how I can use it to solve this problem.

Answer 1

plug both solutions for $s$ in the equation and solve for $b$ and $a$ resp.

Answer 2

If $r$ is the other root. we want $2c+r=0$ and $r(2c)+c^2+d^2=a$ and $(c^2+d^2)r=-b$. The first equation tells us what $r$ is and now it's over.

Remark: We used the fact that if $p$, $q$, and $r$ are the roots, then the coefficient of $s^2$ is $-(p+q+r)$, the coefficient of $s$ is $pq+qr+rp$, and the constant term is $-pqr$.

Answer 3

The polynomial $s^3+as+b$ must be a multiple of $(s-c-di)(s-c+di)=s^2-2cs+(c^2+d^2)$. This implies (with $e$ as third root) $$s^3+as+b=(s-e)(s^2-2cs+(c^2+d^2)\\=s^3-(e+2c)s^2+(2ec+c^2+d^2)s+e(c^2+d^2),$$ hence $e=-2c$ and then $a=2ec+c^2+d^2=-3c^2+d^2$ and $b=e(c^3+d^2)=-2c(c^2+d^2)$.