If $z_1+z_2+z_3=2, z_1^2+z_2^2+z_3^2=3$ and $z_1z_2z_3=4$, find $\frac1{z_1z_2+z_3-1}+\frac1{z_2z_3+z_1-1}+\frac1{z_3z_1+z_2-1}$

Question

Let $z_1, z_2$ and $z_3$ be complex numbers such that $z_1+z_2+z_3=2, z_1^2+z_2^2+z_3^2=3$ and $z_1z_2z_3=4$. Find $\frac1{z_1z_2+z_3-1}+\frac1{z_2z_3+z_1-1}+\frac1{z_3z_1+z_2-1}$

My Attempt:

I got $\sum z_2z_3=\frac12$

Therefore, the cubic equation whose roots are $z_1,z_2,z_3$ is $2x^3-4x^2+x-8=0$. Let this equation be $(1)$.

Also, $\frac1{z_1z_2+z_3-1}=\frac{z_3}{4+z_3^2-z_3}$

Therefore, we need to find the sum of roots of the cubic equation whose roots are of the form $\frac{z_3}{4+z_3^2-z_3}$

Let $x=\frac{z_3}{4+z_3^2-z_3}$

From this, $z_3$ comes out to be $\frac{(x+1)\pm\sqrt{(x+1)^2-16x}}{2}$

But $z_3$ is a root of equation $(1)$. So, maybe I should put the value of $z_3$ here and obtain the cubic in $x$.

But is there any other approach for this question?

I had also tried taking LCM in the required expression but couldn't finish.

Answer 1

Note that: $$\sum_{cyc} \frac{1}{ab+c-1} = \sum_{cyc} \frac{1}{ab+c-(a+b+c-1)} = \sum_{cyc} \frac{1}{(a-1)(b-1)}$$ So we have to evaluate a decently easy value, it becomes:

$$\frac{a+b+c-3}{(a-1)(b-1)(c-1)}$$ where the denominator is easily expanded as a cubic in $-1$. $-1 +1(a+b+c) -1(ab+bc+ca) + abc$

So the final expression should be $\frac{-2}{9}$.

Answer 2

OP found that

1. $z_i$ are the roots of $f(x) = 2x^3 -4x^2 + x - 8 $
2. The sum can be written as $ \sum \frac{z_i}{z_i^2 - z_i + 4 } $.

From there, divide $f(x)$ throughout by $x^2 - x + 4$ and apply partial fractions to show that

$$\frac{ x}{x^2 - x + 4} = \frac{ 2x - 2 } { 9} $$

Hence, $ \sum \frac{ z_i}{z_i^2 - z_i + 4 } = \sum \frac{2z_i-2}{9} = \frac{-2}{9}$.

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More generally, the approach is to

1. Find the polynomial $f(x) = 0 $ for the roots
2. Find an expression for the sum $ = \sum g(x_i)$ involving each root independently
   • This isn't always easily doable. If so, you would need another approach.
3. Find a way to express $g(x_i)$ as a polynomial $h(x_i)$, using the fact that $f(x_i) = 0 $.
   • What we're doing here is essentially finding $ \frac{x}{x^2 - x + 4 } \pmod{2x^3 - 4x^2 + x - 8 } $ in the ring of polynomials, after which we can find the sum through Vieta/Newton identities. It just so happens that the division worked immediately (and might be how they set the question).

Now try finding $ \sum \frac{1}{ z_i^2 - z_i + 4 } $, esp applying the idea of $ \frac{x}{x^2 - x + 4 } \pmod{2x^3 - 4x^2 + x - 8 } $ in the ring of polynomials.