<s>[INST] Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
how can I deduce
$$
\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1}?
$$
I have tried clearing the denominators of the first equation, but that just results in a large mess. I don't know how to continue from there.
 [/INST] I will write $u$ instead of $\omega$, so that $u$ is the primitive third root of unity, and the other one is $u^2=\bar u$.
Below, $v$ will be an element in the list $L=\{1,u,u^2\}$ of all third roots of unity.
The given relation can be rewritten:
$$
2=u\sum\frac 1{a+u}=\sum\frac u{a+u}=\sum\frac 1{1+au^2}
\ ,
$$
where the above sums have $4$ terms each, they are obtained by substituting instead of $a$ formally the values $a,b,c,d$.
It is natural to consider now the polynomial expressions which are the numerators of
$\displaystyle
2-\sum\frac 1{1+av}
$ for $v\in L$. They are
$$
\begin{aligned}
P_v
&:=2(1+av)(1+bv)(1+cv)(1+dv) 
\\
&\qquad\qquad
-\sum \color{gray}{\underbrace{(1+av)}_{\text{omitted}}}(1+bv)(1+cv)(1+dv)
\\
&=2v\; abcd + (bcd+cda+dab+abc) - v(a+b+c+d) -2
\\
\\[3mm]
&\qquad\text{From here we get immediately:}\\[3mm]
vP_v &=
2v^2\; abcd + v(bcd+cda+dab+abc) - v^2(a+b+c+d) -2v
\\
\sum_{v\in L}v\;P_v
&=
2\left(\sum v^2\right)\; abcd + \left(\sum v\right)(bcd+cda+dab+abc)
\\
&\qquad\qquad - \left( \sum v^2\right)(a+b+c+d) -2\left(\sum v\right)
\\
&=0
\ .
\end{aligned}
$$
We have used
$\sum v=\sum v^2=1+u+u^2=0$ for $v$ running in $L$ in the above sums. From $P_u=0$, and its conjugated cousin $P_{u^2}=0$, we obtain $P_1=0$. So:
$$
2=\sum\frac 1{1+a}\ .
$$
$\square$
</s>