Question

If -1,$\alpha$,${\alpha }^3$,${\alpha }^5$,$\overline{\alpha }$,$\overline{{\alpha }^3}$,$\overline{{\alpha }^5}$ are the roots of the equation $z^7$ + 1 = 0. Find the value of $cos\frac{\pi }{7}$ $\times$ $cos\frac{3\pi }{7}$ $\times$ $cos\frac{5\pi }{7}$.

• A. -
• B. -
• C. -
• D. -

Answer 1

The correct option is C

-

Roots of the equation $z^7$ + 1 = 0 are -1,$\alpha$,${\alpha }^3$,${\alpha }^5$,$\overline{\alpha }$,$\overline{{\alpha }^3}$,$\overline{{\alpha }^5}$

$z^7$ = -1

z = $(-1{)}^{\frac{1}{7}}$

= $(cos\pi +isin\pi {)}^{\frac{1}{7}}$

= $cos\frac{\pi +2k\pi }{7}+isin\frac{\pi +2k\pi }{7}$-----------------(1)

Where k = 0,1,2,3,4,5,6

So, $\alpha$ = $cos\frac{\pi }{7}$ + i$sin\frac{\pi }{7}$, $\overline{\alpha }$ = $cos\frac{\pi }{7}$ - i$sin\frac{\pi }{7}$,${\alpha }^3$ = $cos\frac{3\pi }{7}$ + i$sin\frac{3\pi }{7}$,$\overline{{\alpha }^3}$ = $cos\frac{3\pi }{7}$ - i$sin\frac{3\pi }{7}$,${\alpha }^5$ = $cos\frac{5\pi }{7}$ + i$sin\frac{5\pi }{7}$,$\overline{{\alpha }^5}$ = $cos\frac{5\pi }{7}$ - i$sin\frac{5\pi }{7}$.

Given equation can be written as

$(z^7+1)$ = (z + 1) (z - $\alpha$) $(z-(\overline{\alpha }\left)\right)$ $(z-{\alpha }^3)$ $(z-(\overline{{\alpha }^3}\left)\right)$ $(z-{\alpha }^5)$ $(z-(\overline{{\alpha }^5}\left)\right)$

$\frac{z^7+1}{z+1}$ = (z - $\alpha$ ) $(z-(\overline{\alpha }\left)\right)$ $(z-{\alpha }^3)$ $(z-(\overline{{\alpha }^3}\left)\right)$ $(z-{\alpha }^5)$ $(z-(\overline{{\alpha }^5}\left)\right)$

Multiply first with second term,third with fourth term, fifth with sixth term of RHS $\alpha$ & $\overline{\alpha }$ are conjugate each other

When we add two conjugate numbers, we will get 2 $\times$ real part of conjugate number.

$z^6$ - $z^5$ +$z^4$ - $z^3$ + $z^2$ - 2 + 1 = [$z^2$ + 1 - 2$(\alpha -(\overline{\alpha }\left)\right)$] + [$z^2$ + 1 - 2$({\alpha }^3-(\overline{{\alpha }^3}\left)\right)$]

[$z^2$ + 1 - 2$(a^5-(\overline{a^5}\left)\right)$] --------------------(1)

Divide, by $z^3$ on both sides, we get,

$(z^3$ + $\frac{1}{z^3})$ - $(z^2$ + $\frac{1}{z^2})$ + $(z+\frac{1}{z})$ - 1 = $(z+\frac{1}{z}-2cos\frac{\pi }{7}$) $(z+\frac{1}{z}-2cos\frac{3\pi }{7}$) $(z+\frac{1}{z}-2cos\frac{5\pi }{7}$)--------------(1)

Let's assume $z+\frac{1}{z}$ = 2x--------(2)

$(z+\frac{1}{z}{)}^2$ = $z^2$ + $\frac{1}{z^2}$ + 2($z\times \frac{1}{z})$

$z^2$ + $\frac{1}{z^2}$ = $(2x{)}^2$ - 2 = 4$x^2$ - 2 ------------(3)

$(z+\frac{1}{z}{)}^3$ = $z^3$ + $\frac{1}{z^3}$ + 3($z+\frac{1}{z})$

$z^3$ + $\frac{1}{z^3}$ = $(2x{)}^3$ - 3 $\times$ (2x) = 8$x^3$ - 6x-----------(4)

Substitute the value of $(z+\frac{1}{z})$, $z^2$ + $\frac{1}{z^2}$ and $z^3$ + $\frac{1}{z^3}$ in equation (1)

We get,

$(8x^3-6x)-(4x^2-2)+\left(2x\right)-1=(2x-2cos\frac{\pi }{7}$) $(2x-2cos\frac{3\pi }{7}$) $(2x-2cos\frac{5\pi }{7}$)

$8x^3$ - $4x^2$ - 4x + 1 =8 $(x-2cos\frac{\pi }{7}$) $(x-2cos\frac{3\pi }{7}$) $(x-2cos\frac{5\pi }{7}$)

So, $8x^3$ - $4x^2$ - 4x + 1 = 0 has roots $cos\frac{\pi }{7}$.$cos\frac{3\pi }{7}$.$cos\frac{5\pi }{7}$
Product of the roots of above given equation = $cos\frac{\pi }{7}$.$cos\frac{3\pi }{7}$.$cos\frac{5\pi }{7}$ = -$\frac{constantterm}{Coefficientof{x}^3}$ = - $\frac{1}{8}$