# 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.

• Thank you, I have rewritten/updated my post @MathIsNice1729. Apr 1 at 21:56
• Note that $\omega^2 = \overline \omega$ and vice versa. Apr 1 at 22:10
• Thank you @fleablood Apr 1 at 22:18
• If fleablood's tip solves the problem, it would be nice to explain why. Apr 2 at 12:45

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*}

• Well, this is in essence the same story as in my solution, doing very explicit computations... Apr 3 at 16:54
• @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. Apr 4 at 6:37
• 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. Apr 4 at 14:44
• @dan_fulea: Yes, I see your argument. The relationship is rather close. Apr 4 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: $$2=\sum\frac 1{1+a}\ .$$ $$\square$$

• 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? Apr 3 at 19:13
• @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 $\sum v=\sum v^2=0$. Apr 3 at 20:24
• Sorry, I don't see your argument. I think this needs to be shown more detailed. Apr 4 at 6:39
• @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. Apr 4 at 14:36
• Ah, now I see, thanks! Very nice. (+1) Apr 4 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: $$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1} = \\ = \frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3} - (\frac{2}{\omega}+\frac{2}{\overline{\omega}}) =\\= 2 + (\frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3})$$

• The final expression is more complicated than the original one.
– lhf
Apr 4 at 23:56