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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

But try the problem first…

Source

AIME I, 1994, Question 13

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

First hint

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

Second Hint

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

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

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

Final Step

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.