Question

Let x, y, z be complex numbers such that \[x+y+z=2\]\[x^2+y^2+z^2=3\]\[xyz=4\] then the value of \[\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{zx+y-1}=?\]

Answer 1

Options are
a). \(\Large \frac{-2}{9}\)
b).\(\Large \frac{1}{9}\)
c). \(\Large \frac{2}{9}\)
d). \(\Large \frac{-1}{9}\)

Answer 2

@ganeshie8

Answer 3

This is a lot of algebra you have to do, basically try to add all of 3 fractions below, and factor it into the components in terms of the above equation. The first one on the top is going to be xyz(x+y+z), for example. (That is the best thing I see so far.)

Answer 4

First, let's talk about symmetric sums. For a set of \(3\) variables \(\{x,y,z\}\) there are \(3\) possible symmetric sums:
\[\begin{cases}
S_1:=x+y+z\\[1ex]
S_2:=xy+xz+yz\\[1ex]
S_3:=xyz
\end{cases}\]
Why should we care about them? Consider a generic cubic polynomial \(p(t)=t^3+at^2+bt+c\). The fundamental theorem of algebra tells us there are three complex roots \(r_1,r_2,r_3\) that gives a factorization
\[t^3+at^2+bt+c=(t-r_1)(t-r_2)(t-r_3)\]Expanding the right hand side and equating same-power terms, you have
\[\begin{cases}a=-(r_1+r_2+r_3)\\[1ex]
b=r_1r_2+r_1r_3+r_2r_3\\[1ex]
c=-r_1r_2r_3\end{cases}\]These equations are often called Vieta's formulas. You can read some more about them here: https://en.wikipedia.org/wiki/Vieta%27s_formulas

The takeaway here is that the coefficients can be written in terms of the roots, and they coincide with the symmetric sums above such that \(a=-S_1\), \(b=S_2\), and \(c=-S_3\). This means the symmetric sums for the set \(\{r_1,r_2,r_3\}\) uniquely determine the coefficients of \(p(t)\).

Back to your problem: First, notice that
\[x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)\]which means
\[xy+xz+yz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{2^2-3}{2}=\frac{1}{2}\](More generally, any power sum of the form \(x^n+y^n+z^n\) can be written in terms of symmetric sums.)

Then \(\{x,y,z\}\) must be the roots to
\[p(t)=t^3-2t^2+\frac{1}{2}t-4\]which is a pretty good jumping-off point.

Answer 5

The tedious way to continue from there is to explicitly find the roots to \(p(t)\), then reduce the sum you want to find.

The non-tedious way eludes me at the moment.

Answer 6

Got it, the trick is to use the fact that \(x+y+z=2\) to rewrite each denominator. For instance,
\[\begin{align*}
\frac{1}{xy+z-1}&=\frac{1}{xy+(2-x-y)-1}\\[1ex]
&=\frac{1}{xy-x-y+1}\\[1ex]
&=\frac{1}{(x-1)(y-1)}
\end{align*}\]So effectively the sum can be rewritten as
\[\frac{1}{(x-1)(y-1)}+\frac{1}{(y-1)(z-1)}+\frac{1}{(x-1)(z-1)}\]Combine the fractions to get
\[\begin{align*}
\frac{(z-1)+(x-1)+(y-1)}{(x-1)(y-1)(z-1)}&=\frac{x+y+z-3}{(x-1)(y-1)(z-1)}\\[1ex]
&=-\frac{1}{(x-1)(y-1)(z-1)}
\end{align*}\]Expanding the denominator, you get
\[\begin{align*}
(x-1)(y-1)(z-1)&=xyz-xy-xz-yz+x+y+z-1\\[1ex]
&=S_3-S_2+S_1-1\\[1ex]
&=4-\frac{1}{2}+2-1\\[1ex]
&=\frac{9}{2}
\end{align*}\]which means the value of the sum is \(-\dfrac{2}{9}\).

Answer 7

Awesome solution