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# Complex roots and equations | AIME I, 1994 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Complex roots and equations.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Complex roots and equations.

## Complex roots and equations – AIME I, 1994

$x^{10}+(13x-1)^{10}=0$ 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 complex conjugates are taken, find the values of $\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}$

• is 107
• is 850
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Complex Roots

Equation

AIME I, 1994, Question 13

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

here equation gives ${13-\frac{1}{x}}^{10}=(-1)$

$\Rightarrow \omega^{10}=(-1)$ for $\omega=13-\frac{1}{x}$

where $\omega=e^{i(2n\pi+\pi)(\frac{1}{10})}$ for n integer

$\Rightarrow \frac{1}{x}=13- {\omega}$

$\Rightarrow \frac{1}{(x)(\overline{x})}=(13-\omega)(13-\overline{\omega})$

=$170-13(\omega+\overline{\omega})$

adding over all terms $\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}$

=5(170)

=850.