Let θ _ 1 , θ _ 2 , , θ _ 10 be positive valued angles (in radian) such that θ _ 1 + θ _ 2 + ⋯ + θ _ 10 = 2π . Define the complex numbers z _ 1 = e ^ i θ _ 1 , z _ k = z _ k - 1 e ^ i θ _ k for k = 2, 3, , 10 , where i

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Quantrex Academy · Accepted Answer

Correct answer: C. both P and Q are TRUE

Given,& θ _ 1 + θ _ 2 + . . . . . + θ _ 10 = 2π ,~ z _ 1 = e^ i θ _ 1 , z _ k = z _ k - 1 e^ i θ _ k Now,~ z _ k = z _ k - 1 e^ i θ _ k ⇒ z _ 2 = z _ 1 e^ i θ _ 1 , z _ 3 = z _ 2 e^ i θ _ 2 , As,~ z _ 1 = e^ i θ _ 1 ⇒ | z _ 1 | = 1 z _ 2 = z _ 1 e^ iθ ⇒ | z _ 2 | = | z _ 1 | = 1 Similarly~| z _ 1 | = | z _ 2 | = | z _ k | = 1 ⇒ z _ 1 , z _ 2 , , z _ k ~lies on a circle of unit radius. We know,& | z _ k - z _ k - 1 | represents a~line& segment joining z _ k & z _ k - 1 . Both~ z _ k & z _ k - 1 lies on a unit circle Since~ θ _ 1 + θ _ 2 + . . . . + θ _ 10 = 2π We get~ So, | z _ 2 - z _ 1 |, | z _ 3 - z _ 2 | , , | z _ 1 - z _ 10 | are the sides of a decagon circumscribed by a circle~of unit radius We know, the sum of the length of sides& of a decagon is less than the circumference of its circumcircle. ⇒ | z _ 2 - z _ 1 | + | z _ 3 - z _ 2 | + ⋯ + | z _ 10 - z _ 9 | + | z _ 1 - z _ 10 | ≤ 2π Hence, P ~is true. Similarly,& ( z _ 1 ) 2 = e^ i2 θ _ 1 , ( z _ k ) 2 = ( z _ k - 1 ) 2e^ i2 θ _ k , So,~| z _ 2 ^ 2 - z _ 1 ^ 2 |, | z _ 3 ^ 2 - z _ 2 ^ 2 |, | z _ 4 ^ 2 - z _ 3 ^ 2 |, | z _ 5 ^ 2 - z _ 4 ^ 2 |, | z _ 6 ^ 2 - z _ 5 ^ 2 |~are& the& sides of a pentagon circumscribed by a circle of unit radius So,& | z _ 2 ^ 2 - z _ 1 ^ 2 | + | z _ 3 ^ 2 - z _ 2 ^ 2 | + | z _ 6 ^ 2 - z _ 5 ^ 2 | ≤ 2π Similarly,~| z _ 7 ^ 2 - z _ 8 ^ 2 | + + | z _ 1 ^ 2 - z _ 10 ^ 2 | ≤ 2π Adding these~equations, we get | z _ 2 ^ 2 - z _ 1 ^ 2 | + | z _ 3 ^ 2 - z _ 2 ^ 2 | + ⋯ + | z _ 10 ^ 2 - z _ 9 ^ 2 | + | z _ 1 ^ 2 - z _ 10 ^ 2 | ≤ 4π Hence,~ Q & is true.

Let θ _ 1 , θ _ 2 , , θ _ 10 be positive valued angles (in radian)...

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