If α$\alpha$ is a root of the equation x² + x + 1 = 0$x^{2} + x + 1 = 0$ and Σk=1ⁿ(αk + 1αk)² = 20$\sum_{k=1}^{n}\left(\alpha^{k} + \frac{1}{\alpha^{k}}\right)^{2} = 20$, then n$n$ is equal to
Numerical Answer Type:
Enter a numerical valueAnswer: 11 to 11+4 marks
Solution & Explanation
Core Logic
The equation x² + x + 1 = 0$x^2 + x + 1 = 0$ has complex roots which are the non-real cube roots of unity. Thus, we can set α = ω$\alpha = \omega$ (where ω³ = 1$\omega^3 = 1$ and 1 + ω + ω² = 0$1 + \omega + \omega^2 = 0$).
Let's analyze the general term block Tk = (ω^k + (1)/(ω^k))²$T_k = \left(\omega^k + \frac{1}{\omega^k}\right)^2$:
Because ω^k$\omega^k$ is periodic with period 3$3$, let's examine the values of Tk$T_k$ for different values of k$k$:
If k$k$ is a multiple of 3$3$ (k=3m$k=3m$): ω2k = 1, ω^k = 1 Tk = 1 + 1 + 2 = 4$\omega^{2k} = 1, \omega^k = 1 \implies T_k = 1 + 1 + 2 = 4$.
If k$k$ is not a multiple of 3$3$ (k=3m+1$k=3m+1$ or 3m+2$3m+2$): ω2k + ω^k = -1 Tk = -1 + 2 = 1$\omega^{2k} + \omega^k = -1 \implies T_k = -1 + 2 = 1$.
Step 1: Evaluating periodic blocks
Every block of three consecutive terms (k = 1, 2, 3$k = 1, 2, 3$) contributes exactly:
Sum of a block = 1 + 1 + 4 = 6$$\text{Sum of a block} = 1 + 1 + 4 = 6$$
We want the total summation to equal 20$20$. Let's divide 20$20$ by our block value 6$6$:
20 = 3 × 6 + 2$$20 = 3 \times 6 + 2$$
This means the sum must consist of 3$3$ full periodic blocks plus additional terms that add up to 2$2$.
Step 2: Determining the final term count n
The number of terms in 3$3$ full blocks is 3 × 3 = 9$3 \times 3 = 9$ terms, giving a sum of 18$18$.
To get the remaining value of 2$2$, we look at the next terms:
Term 10 (k=10$k=10$, not a multiple of 3) adds 1 Total = 18 + 1 = 19$1 \implies \text{Total} = 18 + 1 = 19$.
Term 11 (k=11$k=11$, not a multiple of 3) adds 1 Total = 19 + 1 = 20$1 \implies \text{Total} = 19 + 1 = 20$.
Hence, the series terminates exactly at n = 11$n = 11$.
Pattern Recognition
Whenever complex roots of unity or cyclic properties show up inside series sums, group terms into blocks based on the underlying period length (3$3$ here) to convert large sums into simple modular arithmetic arithmetic calculations.
Chapter Mix
Class 11 Mathematics: Complex Numbers
Class 11 Mathematics: Sequences and Series
Keywords:#complex roots unity periodic summation#JEE Main 2025 Evening Q71#Complex Numbers JEE Main 2025#Cube Roots of Unity JEE Main 2025
More Complex Numbers Previous-Year Questions
Q17jee_main_2026_21_jan_morningCube Roots of Unity
If x² + x + 1 = 0$x^{2} + x + 1 = 0$ , then the value of (x+ 1x)⁴+(x²+ 1x²)⁴+(x³+ 1x³)⁴+…+(x²⁵+ 1x²⁵)⁴$\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{4}+\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)^{4}+\left(\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}\right)^{4}+\ldots+\left(\mathrm{x}^{25}+\frac{1}{\mathrm{x}^{25}}\right)^{4}$ is :
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$\frac{1}{x} = \frac{1}{\omega} = \omega^2$.
The series is Σk=1²⁵ (ω^k + ω2k)⁴$\sum_{k=1}^{25} (\omega^k + \omega^{2k})^4$.
Evaluate the term Tk = (ω^k + ω2k)⁴$T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of k$k$ with 3.
Powers of x+1/x$x+1/x$ when x$x$ solves x² ± 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
Q19jee_main_2026_21_jan_eveningGeometry of Complex Numbers
Let z$z$ be the complex number satisfying |z-5|≤ 3$|z-5|\leq 3$ and having maximum positive principal argument. Then 34|(5z-12)/(5iz+16)|²$34\left|\frac{5z-12}{5iz+16}\right|^{2}$ is equal to:
A.16$16$
B.12$12$
C.26$26$
D.20$20$
Solution
Related Formula
For maximum argument of z on a circle, the ray from origin is tangent to the circle.$$\text{For maximum argument of } z \text{ on a circle, the ray from origin is tangent to the circle.}$$If |z-a| ≤ r, maximum argument implies θ = (r)/(|a|) and coordinates are x=a ²θ, y=a θ θ$$\text{If } |z-a| \leq r, \text{ maximum argument implies } \sin\theta = \frac{r}{|a|} \text{ and coordinates are } x=a\cos^2\theta, y=a\sin\theta\cos\theta$$
Core Logic
Complex geometry maximum argument diagram for Q19 - JEE Main 2026 Evening
The condition |z-5|≤ 3$|z-5|\leq 3$ represents a disk centered at (5,0)$(5,0)$ with radius 3$3$.
To maximize the principal argument θ$\theta$, the ray from origin must touch the circle in the first quadrant.
The tangent, origin, and center form a right-angled triangle.
Step 1: Locate the Point z
From geometry, hypotenuse c = 5$c = 5$, opposite (radius) r = 3$r = 3$.
The adjacent side (length of tangent) is √(5² - 3²) = 4$\sqrt{5^2 - 3^2} = 4$.
The angle of tangency θ$\theta$ satisfies θ = (3)/(5)$\sin\theta = \frac{3}{5}$ and θ = (4)/(5)$\cos\theta = \frac{4}{5}$.
The point P(z)$P(z)$ lies on the circle and the tangent ray:
For maximum (z)$\arg(z)$ on |z-c| = r$|z-c| = r$ where c$c$ is real, z$z$ coordinates are given by geometric projection: z = √(c²-r²)( θ + i θ)$z = \sqrt{c^2-r^2}(\cos\theta + i\sin\theta)$ where θ = r/c$\sin\theta = r/c$.
Chapter Mix
Class 11 Maths: Complex Numbers
Q21jee_main_2026_22_january_morningCube Roots of Unity
Cube Roots of Unity diagram for Q21 - JEE Main 2026 Morning
Pattern Recognition
Complex expressions with ω$\omega$ raised to large powers almost always exploit rotational symmetry. If coefficients are permuted circularly (A, B, C) → (C, A, B) → (B, C, A)$(A, B, C) \to (C, A, B) \to (B, C, A)$, multiplying by ω$\omega$ maps them to one another, meaning their sum of 3n$3n$ or symmetric powers identically nullifies.
Chapter Mix
Class 11 Maths: Complex Numbers
Q4jee_main_2026_22_january_eveningComplex Equations and Modulus
Let S = z in C : 4z² + z = 0$S = \{z \in \mathbb{C} : 4z^2 + \overline{z} = 0\}$. Then Σz in S |z|²$\sum_{z \in S} |z|^2$ is equal to:
A.(3)/(16)$\frac{3}{16}$
B.(7)/(64)$\frac{7}{64}$
C.(1)/(16)$\frac{1}{16}$
D.(5)/(64)$\frac{5}{64}$
Solution
Related Formula
For z = x + iy$z = x + iy$, z = x - iy$\overline{z} = x - iy$ and |z|² = x² + y²$|z|^2 = x^2 + y^2$.
Core Logic
Substitute z = x + iy$z = x + iy$ into 4z² + z = 0$4z^2 + \overline{z} = 0$:
Separate complex equation into real and imaginary components to systematically find all roots.
Chapter Mix
Class 11 Maths: Complex Numbers
Q17jee_main_2026_23_january_morningGeometry of Complex Numbers
Let S = z : 3 ≤ |2z - 3(1 + i)| ≤ 7$S = \{z : 3 \leq |2z - 3(1 + i)| \leq 7\}$ be a set of complex numbers. Then z in S | ( z + (1)/(2)(5 + 3i) ) |$\min_{z \in S} \left| \left( z + \frac{1}{2}(5 + 3i) \right) \right|$ is equal to:
This represents an annular region bounded by two concentric circles centered at C = (3)/(2) + (3)/(2)i$C = \frac{3}{2} + \frac{3}{2}i$ with radii r₁ = (3)/(2)$r_1 = \frac{3}{2}$ and r₂ = (7)/(2)$r_2 = \frac{7}{2}$.
Geometry of Complex Numbers diagram for Q17 - JEE Main 2026 Morning
Step 1: Identify the Target Point
We need to minimize | z - ( -(5)/(2) - (3)/(2)i ) |$\left| z - \left( -\frac{5}{2} - \frac{3}{2}i \right) \right|$.
Let P$P$ be the point -(5)/(2) - (3)/(2)i$-\frac{5}{2} - \frac{3}{2}i$. The expression represents the distance from point P$P$ to a point z$z$ in the set S$S$.
Step 2: Distance from Center to P
Calculate the distance PC$PC$ between the center of the circles C((3)/(2), (3)/(2))$C\left(\frac{3}{2}, \frac{3}{2}\right)$ and the point P(-(5)/(2), -(3)/(2))$P\left(-\frac{5}{2}, -\frac{3}{2}\right)$:
Since 5 > (7)/(2)$5 > \frac{7}{2}$, the point P$P$ lies outside the outer circle.
Step 3: Minimum Distance Calculation
The shortest distance from an external point to an annular region is the distance to the outer boundary along the line connecting the point to the center.
z in S |z - P| = PC - router$$\min_{z \in S} |z - P| = PC - r_{\text{outer}}$$Minimum Distance = 5 - (7)/(2) = (10 - 7)/(2) = (3)/(2)$$\text{Minimum Distance} = 5 - \frac{7}{2} = \frac{10 - 7}{2} = \frac{3}{2}$$
Pattern Recognition
Transforming |az - b|$|az - b|$ by factoring out a$a$ immediately reveals the true geometric center. Shortest distance to any ring/circle from an external point is always collinear with the center: d - Router$d - R_{outer}$.
Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
More Complex Numbers Questions — jee_main_2025_04_april_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
