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Point Slope Form of a Line

Let α = cis...

Question

Let $α = c i s \frac{2 π}{7}, β = α + α^{2} + α^{4}$ and $γ = α^{3} + α^{5} + α^{6}$ . Then $β$ and $γ$ are the roots of the equation

A

z^{2} + z - 2 = 0

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B

z^{3} + z + 3 = 0

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C

z^{2} + z + 2 = 0

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D

z^{2} + z - 3 = 0

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Solution

The correct option is C $z^{2} + z + 2 = 0$
Given that
$α = c i s \frac{2 π}{7} = cos (\frac{2 π}{7}) + i sin (\frac{2 π}{7})$

Then, by De Moivre's theorem, for a natural number $n$ :

$α^{n} = cos (\frac{2 n π}{7}) + i sin (\frac{2 n π}{7})$

Thus, we have
$β = α + α^{2} + α^{4}$
$= cos (\frac{2 π}{7}) + i sin (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + i sin (\frac{4 π}{7}) + cos (\frac{8 π}{7}) + i sin (\frac{8 π}{7})$
$= cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7}) + i [sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7})]$

Similarly,
$γ = cos (\frac{6 π}{7}) + cos (\frac{10 π}{7}) + cos (\frac{12 π}{7}) + i [sin (\frac{6 π}{7}) + sin (\frac{10 π}{7}) + sin (\frac{12 π}{7})]$

We know that
$cos (2 π - θ) = cos θ$ and $sin (2 π - θ) = - sin θ$

$\Rightarrow cos (\frac{6 π}{7}) = cos (2 π - \frac{8 π}{7}) = cos (\frac{8 π}{7})$
$cos (\frac{10 π}{7}) = cos (2 π - \frac{4 π}{7}) = cos (\frac{4 π}{7})$

$cos (\frac{12 π}{7}) = cos (2 π - \frac{2 π}{7}) = cos (\frac{2 π}{7})$

and
$sin (\frac{6 π}{7}) = sin (2 π - \frac{8 π}{7}) = - sin (\frac{8 π}{7})$
$sin (\frac{10 π}{7}) = sin (2 π - \frac{4 π}{7}) = - sin (\frac{4 π}{7})$
$sin (\frac{12 π}{7}) = sin (2 π - \frac{2 π}{7}) = - sin (\frac{2 π}{7})$

Hence,
$γ = cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7}) - i [sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7})]$

The quadratic equation with $β, γ$ as roots is given by:

$z^{2} - (β + γ) z + β γ = 0$

$β + γ = 2 [cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7})]$

Consider the sum $cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7})$

This is of the form $cos A + cos B + cos C$ with $A + B + C = 2 π$ .
$cos A + cos B + cos C = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2}) + cos (2 π - (A + B))$

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

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

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

$= 2 cos (π - \frac{C}{2}) [2 cos (\frac{A}{2}) cos (\frac{B}{2})] - 1$

$= - 4 cos (\frac{C}{2}) cos (\frac{A}{2}) cos (\frac{B}{2}) - 1$

So,
$β + γ = 2 [cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7})]$
$= 2 [- 1 - 4 cos (\frac{π}{7}) cos (\frac{2 π}{7}) cos (\frac{4 π}{7})]$
$= 2 [- 1 - 4 (\frac{- 1}{8})] = - 1$

$β γ = {[cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7})]^{2} + [sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7})]}^{2}$

$= 3 + 2 {cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) + cos (\frac{8 π}{7}) cos (\frac{4 π}{7}) + cos (\frac{2 π}{7}) cos (\frac{8 π}{7}) + . . .$ $+ sin (\frac{2 π}{7}) sin (\frac{8 π}{7})}$
$= 3 + 2 {[cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) + sin (\frac{2 π}{7}) sin (\frac{4 π}{7})] + [cos (\frac{8 π}{7}) cos (\frac{4 π}{7}) + . . .$ $+ sin (\frac{2 π}{7}) sin (\frac{8 π}{7})]}$

Since $cos A cos B + sin A sin B = cos (A - B)$ ,

$β γ = 3 + 2 {[cos (\frac{2 π}{7})] + [cos (\frac{4 π}{7})] + [cos (\frac{6 π}{7})]}$
$= 3 + 2 {[cos (\frac{2 π}{7})] + [cos (\frac{4 π}{7})] + [cos (\frac{8 π}{7})]}$

As shown above

$[cos (\frac{2 π}{7})] + [cos (\frac{4 π}{7})] + [cos (\frac{8 π}{7})] = - 1 - 4 cos (\frac{π}{7}) cos (\frac{2 π}{7}) cos (\frac{4 π}{7})$

$∴ β γ = 3 + 2 {- 1 - 4 cos (\frac{π}{7}) cos (\frac{2 π}{7}) cos (\frac{4 π}{7})}$
$= 3 + 2 {- 1 - 4 (\frac{- 1}{8})}$
$= 2$

The equation for which $β$ and $γ$ are the roots:
$z^{2} - (β + γ) z + β γ = 0$

$z^{2} + z + 2 = 0$

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