Roots and coefficients

Congratulations Sue Liu of Madras College, St Andrew's on your solution to this problem. The title of this problem is the clue to getting a neat solution. We are given: Just consider the cubic equation $$z^3+a{z}^2+bz+c=(z-z_1)(z-z_2)(z-z_3)=0$$ with roots $z_1, z_2$ and $z_3$. We know that $a= -(z_1+z_2+z_3)$, $b= z_1z_2+z_2z_3+z_3z_1$ and $c= -(z_1z_2z_3)$. As we are given the product of the roots in (1) we know that $c= -1$.

A little experimentation with the second identity (2) gives a relationship between $a$ and $b$.

From (2) $$\frac{z_1{z}_2+z_2{z}_3+z_3{z}_1}{z_1{z}_2{z}_3}=x$$ and, using (1) this gives $$z_1{z}_2+z_2{z}_3+z_3{z}_1=x$$ Hence the cubic equation is $$z^3-x{z}^2+xz-1=(z-1)(z^2+(1-x)z+1)=0(3).$$ The factor $(z - 1)$ of this cubic equation shows that one of the values of z must be 1.

For the other two roots to be real the quadratic factor in (3), $$z^2+(1-x)z+1=0$$ must have real roots in which case If $(x + 1)(x - 3)\geq 0$ then $x \leq -1$ or $x\geq 3$. So the other two roots are real when the value of $$z_1+z_2+z_3=\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=x$$ is less than or equal to -1 or greater than or equal to 3.