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 $\boxed{850}$ .

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1994 AIME Problems/Problem 13

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