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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:
\begin{eqnarray} \\z_1z_2z_3 &=& 1 \quad &(1) \\ \\z_1+z_2+z_3 &=& {1\over z_1} + {1\over z_2} + {1\over z_3} = x \quad &(2).\\ \end{eqnarray}
Just consider the cubic equation $$z^3 + az^2 + bz +c = (z - z_1)(z - z_2)(z - z_3) = 0$$ 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) $${{z_1z_2+z_2z_3+z_3z_1}\over {z_1z_2z_3}} = x$$ and, using (1) this gives $${z_1z_2 + z_2z_3 + z_3z_1}= x$$ Hence the cubic equation is $$z^3 - xz^2 + xz - 1 = (z - 1)(z^2 + (1 - x)z + 1) = 0 \quad (3).$$ 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
\begin{eqnarray} (1 - x)^2 - 4 &\geq& 0 \\ x^2 - 2x - 3 &\geq& 0 \\ (x + 1)(x - 3) &\geq& 0.\\ \end{eqnarray}
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 $$ z_1 + z_2 + z_3 = {1\over z_1} + {1\over z_2} + {1\over z_3}= x $$ is less than or equal to -1 or greater than or equal to 3.