# Difference between revisions of "1994 AIME Problems/Problem 13"

## Problem

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 $\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}.$

## Solution

Let $t = 1/x$. After multiplying the equation by $t^{10}$, $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$.

Using DeMoivre, $13 - t = e^\frac {(2k + 1)\pi}{10}$ where $k$ is an integer between $0$ and $9$. $t = 13 - e^\frac {(2k + 1)\pi}{10} \Rightarrow \bar{t} = 13 - e^{ - \frac {(2k + 1)\pi}{10}}$.

Since $e^{iy} + e^{ - iy} = 2\cos y$, $t\bar{t} = 170 - 2\cos \frac {(2k + 1)\pi}{10}$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\pi$ are involved in the product..

The expression to find is $\sum t\bar{t} = 850 - 2\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$.

But $\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0$ so the sum is $\boxed{850}$.