Question

let v and w be distinct, randomly chosen roots (real or complex ) of the equation z
${z}^{9} - 1 = 0$
the probability that
$1 \leqslant |v + w|$
is?

Answer 1

is the answer option (b) ???
3/c(9,2).

Answer 2

Concept: Simply put, probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

Given: v and w be distinct, randomly chosen roots (real or complex ) of the equation z, $z^9 - 1 = 0$

To find:  the probability that 1 ≤ | v + w |

Solution:

$z^9 - 1 = 0\\\\z^9 = 1$

z = r(cos θ + i sin θ)

z = cos θ + i sin θ

v = cos α + i sin α

w = cos β + i sin β

1 ≤ | cos α + i sin α + cos β + i sin β |

1 ≤ |(cos α + cos β) + i (sin α + sin β) |

z = x + iy

$|z| = \sqrt{x^2 + y^2 }$

1 ≤ $\sqrt{(cos \alpha + cos\beta )^2 + (sin\alpha + sin \beta )^2}$

1 ≤ $cos^2\alpha + cos^2\beta +2cos\alpha cos\beta + sin ^2\alpha + sin^2\beta + 2sin\alpha sin\beta$

As $cos^2\alpha + sin^2\alpha =1$

1 ≤ 1 + 1 + 2 (cos α cos β + sin α sin β)

$\frac{-1}{2} \leq cos(\alpha -\beta )$

α - β = [0, $\frac{2\pi }{3}$]

α - β = [$\frac{4\pi }{3} , 2\pi ]$

α - β = [$\frac{-2\pi }{3} , 0]$

$\frac{360}{9} = 4$    [angle between two roots]

Favourable cases = 9*6=54

Total cases = 2!$9\\ C_2$ = $\frac{9*8*2}{2} = 72$

therefore probability = $\frac{54}{72} = 1.33$

Hence, the probability that 1 ≤ | v + w | is 1.33.

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