Question

The equation $x^{10}+(13x-1)^{10}=0$ has $10$ complex roots $r_1,$ $\overline{r_1},$ $r_2,$ $\overline{r_2},$ $r_3,$ $\overline{r_3},$ $r_4,$ $\overline{r_4},$ $r_5,$ $\overline{r_5},$ where the bar denotes complex conjugation. Find the value of $\begin{aligned}&\frac1{r_1\overline{r_1}}+\frac1{r_2\overline{r_2}}+\frac1{r_3\overline{r_3}}\\&\quad+\frac1{r_4\overline{r_4}}+\frac1{r_5\overline{r_5}}.\end{aligned}$

Answer

Answer: 850

Concepts:roots of unity complex number summation

Difficulty rating: 2650

Small Hint:

Set $\frac{x}{13x-1}=\zeta,$ where $\zeta^{10}=-1$

Big Hint:

Express $\frac{1}{|x|^2}$ in terms of $\zeta+\overline\zeta$ and sum over the five conjugate pairs

Solution:

Let $\zeta=\frac{x}{13x-1},$ so $\zeta^{10}=-1$ and $\begin{aligned}x&=\frac{\zeta}{13\zeta-1},\\\frac1x&=13-\zeta^{-1}.\end{aligned}$ Since $|\zeta|=1,$ $\begin{aligned}\frac1{|x|^2}&=|13-\zeta^{-1}|^2\\&=170-13(\zeta+\overline\zeta).\end{aligned}$ Summing one value for each of the five conjugate pairs gives $5\cdot170$ minus $13$ times the sum of all ten roots of $z^{10}+1,$ which is $0.$ The requested value is $850.$

1994 AIME Problem 13

Answer: 850

Difficulty rating: 2650

Small Hint:

Big Hint:

Solution:

Problem 13 in Other Years