If z ^ 2 + z + 1 = 0, z ∈ C , then | ∑ _ n = 1 ^ 15 ( z ^ n + ( - 1 ) ^ a 1 z ^ n ) ^ 2 | is equal to _____.

Correct answer: 0 Given, z ^ 2 + z + 1 = 0 ⇒ z = w, w ^ 2 Now, | ∑ _ n = 1 ^ 15 ( z ^ n + (- 1) 1 z ^ n ) ^ 2 | = | ∑ _ n = 1 ^ 15 ( z ^ 2n + 1 z ^ 2n + 2 (- 1) ^ n )| = | ∑ _ n = 1 ^ 15 w ^ 2n + 1 w ^ 2n + 2 (- 1) ^ n | = | w ^ 2 (1 - w ^ 30 ) 1 - w ^ 2 + 1 w ^ 2 (1 - 1 w ^ 30 ) 1 - 1 w ^ 2 + 2(- 1)| = | w ^ 2 (1 - 1) 1 - w ^ 2 + 1 w ^ 2 (1 - 1) 1 - 1 w ^ 2 - 2| = |0 + 0 - 2| = 2

JEE Main 202226 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 26 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2022 · 26 June · Evening Shift · Mathematics · Complex Number

If z ^ 2 + z + 1 = 0, z ∈ C , then | ∑ _ n = 1 ^ 15 ( z ^ n + ( - 1...

{z}^{2} + z + 1 = 0, z \in C

, then \left|\mathop{∑}_{n = 1}^{15}{\left({z}^{n} + ( - 1{)}^{a}\dfrac{1}{{z}^{n}}\right)

}^{2}\right|

is equal to _____.

Correct answer

Step-by-step solution

Given, ${z}^{2} + z + 1 = 0 ⇒ z = w, {w}^{2}$

Now, \left|\mathop{∑}_{n = 1}^{15}{\left({z}^{n} + \left(- 1\right)\dfrac{1}{{z}^{n}}\right)}^{2}\right| = \left|\mathop{∑}_{n = 1}^{15}\left({z}^{2n} + \dfrac{1}{{z}^{2n}} + 2{\left(- 1\right)}^{n}\right)\right| $$ = \left|\mathop{∑}_{n = 1}^{15}{w}^{2n} + \dfrac{1}{{w}^{2n}} + 2{\left(- 1\right)}^{n}\right| $$ = \left|\dfrac{{w}^{2}\left(1 - {w}^{30}\right)}{1 - {w}^{2}} + \dfrac{\dfrac{1}{{w}^{2}}\left(1 - \dfrac{1}{{w}^{30}}\right)}{1 - \dfrac{1}{{w}^{2}}} + 2\left(- 1\right)\right| $$ = \left|\dfrac{{w}^{2}\left(1 - 1\right)}{1 - {w}^{2}} + \dfrac{\dfrac{1}{{w}^{2}}\left(1 - 1\right)}{1 - \dfrac{1}{{w}^{2}}} - 2\right| $$ = \left|0 + 0 - 2\right| = 2$

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