AIME I Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

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

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


Complex Roots


Check the Answer

Answer: is 850.

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

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

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


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



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