Roots and Coefficients
If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of
these numbers must be 1. Now for the complexity! When are the other
numbers real and when are they complex?
Age
16 to 18
Challenge level
If \[z_1 z_2 z_3 = 1\] and \[z_1 + z_2 + z_3 = \frac{1}{z_1} + \frac{1}{z_2} +\frac{1}{z_3}\] then show that at least one of these numbers must be 1.
Now for the complexity! When are the other numbers real and when are they complex?
The hint is in the title here!
If $z_1z_2z_3=1$ what can you say about ${1\over z_1} + {1\over z_2} + {1\over z_3}$?
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.