हिंदी

Let z = 1-i32, i = -1. Then the value of 21+(z+1z)3+(z2+1z2)+(z3+1z3)3+......+(z21+1z21)3 is ______. -

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प्रश्न

Let z = `(1 - isqrt(3))/2`, i = `sqrt(-1)`. Then the value of `21 + (z + 1/z)^3 + (z^2 + 1/z^2) + (z^3 + 1/z^3)^3 + ...... + (z^21 + 1/z^21)^3` is ______.

विकल्प

  • 11

  • 12

  • 13

  • 14

MCQ
रिक्त स्थान भरें

उत्तर

Let z = `(1 - isqrt(3))/2`, i = `sqrt(-1)`. Then the value of `21 + (z + 1/z)^3 + (z^2 + 1/z^2) + (z^3 + 1/z^3)^3 + ...... + (z^21 + 1/z^21)^3` is 13.

Explanation:

Given, –z = `(-1 + isqrt(3))/2` and i = `sqrt(-1)`

Let –z = ω or z = ω

Now, `z + 1/z = -ω - 1/ω = (-ω^2 - 1)/ω = ω/ω` = 1  ...(i)

And, `z^2 + 1/z^2 = (-ω)^2 + 1/(-ω)^2`

`ω^2 + 1/ω^2 = (ω^4 + 1)/ω^2 = (ω + 1)/ω^2 = (-ω^2)/ω^2` = –1  ...(ii)

Again, `z^3 + 1/z^3 = (-ω)^3 + 1/(-ω)^3` = –1 – 1 = –2  ...(iii)

From equation (i), (ii) and (iii)

`(z + 1/z)^3 + (z^2 + 1/z^2)^3 + (z^3 + 1/z^3)^3`

= 1 – 1 – 8 = –8  ...(iv)

Again, `z^6 + 1/z^6 = (-ω)^6 + 1/(-ω)^6` = (1 + 1)3 = 8

`(z^3 + 1/z^3)^3 + (z^6 + 1/z^6)^3` = 0  ...(v)

`(z^9 + 1/z^9)^3 + (z^12 + 1/z^12)^3` = 0  ...(vi)

`(z^15 + 1/z^15)^3 + (z^18 + 1/z^18)^3` = 0  ...(vii)

`z^21 + 1/z^21` = –8  ...(viii)

According to the question, by using the above equations

⇒ 21 – 8 = 13

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