monkeyman130yb

2022-03-30

$\alpha$ and $\beta$ are the roots of the quadratic equation.

$\omega \ne 1$ and ${\omega}^{13}=1$

If $\alpha =\omega +{\omega}^{3}+{\omega}^{4}+{\omega}^{-4}+{\omega}^{-3}+{\omega}^{-1}$

$\beta ={\omega}^{2}+{\omega}^{5}+{\omega}^{6}+{\omega}^{-6}+{\omega}^{-5}+{\omega}^{-2}$,

then quadratic equation, whose roots are $\alpha$ and $\beta$ is:

I tried finding the sum and product of roots and placing in the equation ${x}^{2}-(\alpha +\beta )x+(\alpha \times \beta )=0$ but I was not able to solve for it.

Lana Hamilton

Beginner2022-03-31Added 12 answers

Step 1

By the formula for a finite geometric series,

$\alpha +\beta =\sum _{i=-6}^{-1}{\omega}^{i}+\sum _{i=1}^{6}{\omega}^{i}$

$=\sum _{i=7}^{12}{\omega}^{i}+\sum _{i=1}^{6}{\omega}^{i}$

$=\frac{(1-{\omega}^{13})}{1-\omega}-1$= $-1({\omega}^{13}=1)$

$-1=-1$

For the product $\alpha \beta$, note that ${\omega}^{n}={\omega}^{-n}$

Taking $\omega$ as ${e}^{2i\frac{\pi}{13}}=\mathrm{cos}\left(2\frac{\pi}{13}\right)+i\mathrm{sin}\left(2\frac{\pi}{13}\right)$ for example, $\alpha =2(\mathrm{cos}\left(2\frac{\pi}{13}\right)+\mathrm{cos}\left(6\frac{\pi}{13}\right)+\mathrm{cos}\left(8\frac{\pi}{13}\right))$ and you can do the same for $\beta$. Then you can use the product-to-sum formula

$\mathrm{cos}A\mathrm{cos}B=\frac{1}{2}(\mathrm{cos}(A-B)+\mathrm{cos}(A+B))$

by expanding the brackets, there are nine terms in this form.

Tristatex9tw

Beginner2022-04-01Added 18 answers

Step 1

We have

$1+\alpha +\beta =1+\omega +{\omega}^{2}+\cdots +{\omega}^{12}=0$

and so$\alpha +\beta =-1$

Therefore,$\alpha \beta =-(\alpha +{\alpha}^{2})=-3$

Unfortunately, I can't see shortcut to compute$\alpha +{\alpha}^{2}$ except by expanding it.

We have

and so

Therefore,

Unfortunately, I can't see shortcut to compute

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