Advanced complex numbers/roots of unity

Question

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.

Answer 1

A calculation by foot. We obtain from \begin{align*} \color{blue}{0}&\color{blue}{=\frac{1}{a+\omega}+\frac{1}{b+\omega} +\frac{1}{c+\omega}+\frac{1}{d+\omega}-\frac{2}{\omega}}\tag{1}\\ \end{align*} by multiplication with the common denominator $(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega$: \begin{align*} \color{blue}{0}&=(b+\omega)(c+\omega)(d+\omega)\omega\\ &\qquad+(a+\omega)(c+\omega)(d+\omega)\omega\\ &\qquad+(a+\omega)(b+\omega)(d+\omega)\omega\\ &\qquad+(a+\omega)(b+\omega)(c+\omega)\omega\\ &\qquad-2(a+\omega)(b+\omega)(c+\omega)(d+\omega)\\ &=bcd\omega+(bc+bd+cd)\omega^2+(b+c+d)\omega^3+\omega^4\\ &\qquad+acd\omega+(ac+ad+cd)\omega^2+(a+c+d)\omega^3+\omega^4\\ &\qquad+abd\omega+(ab+ad+bd)\omega^2+(a+b+d)\omega^3+\omega^4\\ &\qquad+abc\omega+(ab+ac+bc)\omega^2+(a+b+c)\omega^3+\omega^4\\ &\qquad-2abcd\\ &\qquad-2(abc+abd+acd+bcd)\omega\\ &\qquad-2(ab+ac+ad+bc+bd+cd)\omega^2\\ &\qquad-2(a+b+c+d)\omega^3\\ &\qquad-2\omega^4\\ &\,\,\color{blue}{=\left(2-(abc+abd+acd+bcd)\right)\omega-2abcd+a+b+c+d}\tag{2} \end{align*} In the last step (2) we observe that terms with $\omega^2$ cancel away and we also use the identities \begin{align*} \omega^3=1,\quad\omega^4=\omega \end{align*}

We note from (1) and (2) we can write (1) as \begin{align*} \color{blue}{0=\frac{A\omega+B}{(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega}}\tag{3} \end{align*} and since $A,B\in\mathbb{R}$ and $\omega\in\mathbb{C}\setminus{\mathbb{R}}$ we conclude $A=B=0$, so that \begin{align*} \color{blue}{abc+abd+acd+bcd}&\color{blue}{=2}\tag{4}\\ \color{blue}{a+b+c+d}&\color{blue}{=2abcd}\\ \end{align*} follows.

On the other hand we consider the expression \begin{align*} \color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}}&\color{blue}{=z} \end{align*}

Multiplication of the LHS with the common denominator $(1+a)(1+b)(1+c)(1+c)$ gives

\begin{align*} &(b+1)(c+1)(c+1)+(a+1)(c+1)(d+1)\\ &\qquad\quad+(a+1)(b+1)(d+1)+(a+1)(b+1)(c+1)\\ &\quad=1+(b+c+d)+(bc+bd+cd)+bcd\\ &\qquad\quad+1+(a+c+d)+(ac+ad+cd)+acd\\ &\qquad\quad+1+(a+b+d)+(ab+ad+bd)+abd\\ &\qquad\quad+1+(a+b+c)+(ab+ac+bc)+abc\\ &\quad=4+3(a+b+c+d)\\ &\qquad\quad+2(ab+ac+ad+bc+bd+cd)\\ &\qquad\quad+abc+abd+acd+bcd\\ &\,\,\color{blue}{=2\left(3+3abcd+ab+ac+ad+bc+bd+cd\right)}\tag{5} \end{align*} In the last line (5) we used the identities from (4).

Similarly, multiplication of the RHS with the common denominator gives

\begin{align*} &z(a+1)(b+1)(c+1)(d+1)\\ &\qquad=z(1+(a+b+c+d)\\ &\qquad\quad+(ab+ac+ad+bc+bd+cd)\\ &\qquad\quad+(abc+abd+acd+bcd)+abcd)\\ &\,\,\color{blue}{\qquad=z(3+3abcd+ab+ac+ad+bc+bd+cd)}\tag{6} \end{align*} Again in the last line (6) we used the identities from (4) for simplification.

Comparing (5) and (6) we conclude \begin{align*} \color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2} \end{align*}

Answer 2

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: $$ 2=\sum\frac 1{1+a}\ . $$ $\square$

Answer 3

The keys is Vieta's formulas.

Let $u$ be a root of $x^3 = 1 $. Then $u+a$ is a root of $(x-a)^3 = 1$ and $\frac{1}{u+a}$ is a root of $(\frac{1}{x}-a)^3=1 $. Multiplying by $x^3$ and expanding the coefficients we have $$(1+a^3) x^3 -3a^2x^2 + 3ax - 1 = 0$$ Using Vieta's formulas we get that $\frac{3a^2}{1+a^3}$ is equal to $$ \frac{3a^2}{1+a^3} = \sum_{u^3 =1} \frac{1}{(u+a)} \tag{1} $$ We can apply this to our equation. We know that $$\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega}$$ and by conjugating: $$\frac{1}{a + \overline{\omega}} + \frac{1}{b +\overline{\omega}} + \frac{1}{c +\overline{\omega}} + \frac{1}{d + \overline{\omega}} = \frac{2}{\overline{\omega}}$$ Now $\omega$ and $\overline{\omega}$ are two of the three roots of unity (solutions to $x^3 = 1$), the other being $1$. From this we conclude: $$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1} = \\ = \frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3} - (\frac{2}{\omega}+\frac{2}{\overline{\omega}}) =\\= 2 + (\frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3}) $$