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.

  • 4
    $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. $\endgroup$ Commented Apr 1, 2021 at 21:50
  • 2
    $\begingroup$ Thank you, I have rewritten/updated my post @MathIsNice1729. $\endgroup$
    – dgeyfman
    Commented Apr 1, 2021 at 21:56
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    $\begingroup$ Note that $\omega^2 = \overline \omega$ and vice versa. $\endgroup$
    – fleablood
    Commented Apr 1, 2021 at 22:10
  • $\begingroup$ Thank you @fleablood $\endgroup$
    – dgeyfman
    Commented Apr 1, 2021 at 22:18
  • $\begingroup$ If fleablood's tip solves the problem, it would be nice to explain why. $\endgroup$
    – Rob Arthan
    Commented Apr 2, 2021 at 12:45

3 Answers 3


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*}

  • $\begingroup$ Well, this is in essence the same story as in my solution, doing very explicit computations... $\endgroup$
    – dan_fulea
    Commented Apr 3, 2021 at 16:54
  • $\begingroup$ @dan_fulea: I don't think so. The essence in my answer are the relations (4) and their explicit usage in (5) and (6). This is not used in your answer. $\endgroup$ Commented Apr 4, 2021 at 6:37
  • $\begingroup$ Your relation $(4)$ is equivalent to my relation $P_\omega=0$ (and/or its conjugated form), just isolate the pieces w.r.t. the basis $1,\omega$ of $\Bbb Q(\omega)$ over $\Bbb Q$. Then you do not really need $(5)$ and $(6)$ separatedly, but rather $(5)-(6)$ for the special value of $z$, namely $z=2$, that makes it coincide with $(4)$. This is equivalent to / well, at least implied from the fact that $P_1$ is a linear combination of $P_\omega$, $P_{\omega^2}$ - in my notations. $\endgroup$
    – dan_fulea
    Commented Apr 4, 2021 at 14:44
  • $\begingroup$ @dan_fulea: Yes, I see your argument. The relationship is rather close. $\endgroup$ Commented Apr 4, 2021 at 14:56

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

  • $\begingroup$ A very nice approach and it looks promising. There's just one aspect where I miss presumably something obvious. I see $P_u=P_{u^2}=0$. But I don't see the validity of $0=\sum_{v\in L}v P_v$. Would you mind providing some more info on that? $\endgroup$ Commented Apr 3, 2021 at 19:13
  • $\begingroup$ @MarkusScheuer In the formula of $vP_v$ there appear only expressions in $v$ and in $v^2$. (There is no "real constant" part.) Now use $\sum v=\sum v^2=0$. $\endgroup$
    – dan_fulea
    Commented Apr 3, 2021 at 20:24
  • $\begingroup$ Sorry, I don't see your argument. I think this needs to be shown more detailed. $\endgroup$ Commented Apr 4, 2021 at 6:39
  • $\begingroup$ @MarkusScheuer I edited so that the formula for $vP_v$ is explicitly displayed. Then the sum of $vP_v$ is also explicitly shown. The computational idea is simple, we have a linear relation with coefficients in $\Bbb Q(u)=\Bbb Q(\omega)$ among the polynomials $P_1$, $P_u$, $P_{u^2}$. In your solution, the real and imaginary parts are isolated - as a matter of taste. $\endgroup$
    – dan_fulea
    Commented Apr 4, 2021 at 14:36
  • $\begingroup$ Ah, now I see, thanks! Very nice. (+1) $\endgroup$ Commented Apr 4, 2021 at 14:53

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: $$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1} = \\ = \frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3} - (\frac{2}{\omega}+\frac{2}{\overline{\omega}}) =\\= 2 + (\frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3}) $$

  • $\begingroup$ The final expression is more complicated than the original one. $\endgroup$
    – lhf
    Commented Apr 4, 2021 at 23:56

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