Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Complex roots and equations.
\(x^{10}+(13x-1)^{10}=0\) 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 \(\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}\)
Integers
Complex Roots
Equation
Answer: is 850.
AIME I, 1994, Question 13
Complex Numbers from A to Z by Titu Andreescue
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.
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad