If x2 + x + 1 = 0 , then the value of (x+(1/x))4+(x2+(1/x)2)4+(x3+(1/x)3)4+ldots+(x25+(1/x)25)4 is :

Solution & Explanation

Related Formula

x² + x + 1 = 0 \Rightarrow x = ω, ω²

Properties of cube roots of unity: $ω³ = 1$ $\omega^3 = 1$ and $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$.

Core Logic

Let $α = ω$ $\alpha = \omega$. Then $(1)/(x) = (1)/(ω) = ω²$ . The series is $Σk=1²⁵ (ω^k + ω2k)⁴$ . Evaluate the term $Tk = (ω^k + ω2k)⁴$ $T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of $k$ $k$ with 3.

Step 1: Cyclic Evaluation

Case 1: $k = 3m$ $k = 3m$ (multiples of 3)

T3m = (ω3m + ω6m)⁴ = (1 + 1)⁴ = 2⁴ = 16

There are $8$ $8$ such multiples up to 25 ( $3, 6, 9, , 24$ $3, 6, 9, \dots, 24$).

Case 2: $k \neq 3m$ $k \neq 3m$ (non-multiples of 3) For $k = 1, 2, 4, 5,$ $k = 1, 2, 4, 5, \dots$ $ω^k + ω2k$ $\omega^k + \omega^{2k}$ will always be $ω + ω²$ $\omega + \omega^2$ or $ω² + ω$ $\omega^2 + \omega$. Since $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$, we have $ω + ω² = -1$ $\omega + \omega^2 = -1$.

Tk \neq 3m = (-1)⁴ = 1

There are $25 - 8 = 17$ $25 - 8 = 17$ such non-multiples up to 25.

Step 2: Total Sum

Sum = 17 \times 1 + 8 \times 16

= 17 + 128 = 145

Pattern Recognition

Powers of $x+1/x$ $x+1/x$ when $x$ $x$ solves $x² \pm x + 1 = 0$ $x^2 \pm x + 1 = 0$ perfectly orbit around periods of 3 or 6. Isolate the $3m$ $3m$ resonant beats (which hit pure scalars like $1+1=2$ $1+1=2$) versus the out-of-phase beats (which collapse to $-1$ $-1$ or $1$ $1$ via basic $ω$ $\omega$ identities).

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 9

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument

α

denotes the number of solutions of

|1 - i|^x = 2^x

and

β = ((|z|)/( (z)))

, where

z = (π)/(4) (1 + i)⁴ ( 1 - \sqrt(π) i\sqrt(π) + i + \sqrt(π) - i1 + \sqrt(π) i), i = \sqrt(-1)

, then the distance of the point

(α, β)

from the line

4x - 3y = 7

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

|1 - i|^x = 2^x (\sqrt(2))^x = 2^x 2x/2 = 2^x

This implies $(x)/(2) = x x = 0$ $\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so $α = 1$ $\alpha = 1$.

Step 1: Simplify complex number z

(1+i)⁴ = ((1+i)²)² = (1 + i² + 2i)² = (2i)² = -4

Thus, $z = -π ( (1-\sqrt(π)i)(\sqrt(π)-i)π + 1 + (\sqrt(π)-i)(1-\sqrt(π)i)1 + π )$

Step 2: Simplify Bracket

Let's expand the terms directly: $z = (π)/(4)(-4) [ \sqrt(π) - π i - i - \sqrt(π)π + 1 + \sqrt(π) - i - π i - \sqrt(π)1 + π ]$

= -π [ (-i(π+1))/(π+1) + (-i(π+1))/(π+1) ]

= -π [ -i - i ] = 2π i

Step 3: Find beta

For $z = 2π i$ $z = 2\pi i$: $|z| = 2π$ $|z| = 2\pi$ and $(z) = (π)/(2)$ $\arg(z) = \frac{\pi}{2}$.

β = (|z|)/( (z)) = (2π)/(π/2) = 4

Step 4: Distance from Line

Distance of point $(α, β) = (1, 4)$ $(\alpha, \beta) = (1, 4)$ from the line $4x - 3y - 7 = 0$ $4x - 3y - 7 = 0$:

D = |4(1) - 3(4) - 7|\sqrt(4² + (-3)²) = (|4 - 12 - 7|)/(5) = (|-15|)/(5) = 3

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

If $x² + x + 1 = 0$ $x^{2} + x + 1 = 0$ , then the value of $(x+ 1x)⁴+(x²+ 1x²)⁴+(x³+ 1x³)⁴+\dots+(x²⁵+ 1x²⁵)⁴$ is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Cyclic Evaluation

Step 2: Total Sum

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers and Quadratic Equations Questions — jee_main_2026_21_jan_morning

Select Your Target Exam

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Solution & Explanation

Related Formula

Core Logic

Step 1: Cyclic Evaluation

Step 2: Total Sum

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers and Quadratic Equations Questions — jee_main_2026_21_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

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