# If $a =\cos (\frac{2 \pi}{7})+i \sin (\frac{2 \pi}{7})$ then construct a quadratic equation.

If $a =\cos (\frac{2 \pi}{7})+i \sin (\frac{2 \pi}{7})$, then find a quadratic equation whose roots are $\alpha = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^7$ .

Using the fact that sum of $7th$ roots of unity is $0$, I obtained sum of roots as

$\alpha+\beta=-a^6$ and product of roots=$\alpha \beta=1+a^9-a^{10}+a^{11}$ which still quite complicated.

Could some help me as how can I proceed further?

• @Travis $-a$ or $-1/a$? – Akira May 26 '16 at 20:02
• Oops, you're right of course, $a^6 = a^{-1}$. – Travis Willse May 26 '16 at 20:12
• Should $\beta$ be $a^3 + a^5 + a^6$? That would make the two roots complex conjugates, and hence make the resulting quadratic real. – Travis Willse May 26 '16 at 20:12

## 1 Answer

There is an obvious typo in the question $\beta$ should be $a^3+a^5+a^6$. Since $a$ is a complex 7th root of unity we have $1+a+a^2+a^3+a^4+a^5+a^6=0$, so $\alpha+\beta=-1$.

Multiplying out and using $a^7=1$ we have $\alpha\beta=3+a+a^2+a^3+a^4+a^5+a^6=2$. So $\alpha,\beta$ are the roots of the quadratic $x^2-x+2=0$.