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
Convert standard coordinate complex numbers into Euler form immediately when large powers are present.
Chapter Mix
Class 11 Maths: Complex Numbers
Q9jee_main_2026_24_january_morningLocus in Complex Plane
Let S = z in C : | (z - 6i)/(z - 2i) | = 1 and | (z - 8 + 2i)/(z + 2i) | = (3)/(5)$S = \left\{ z \in \mathbb{C} : \left| \frac{z - 6i}{z - 2i} \right| = 1 \text{ and } \left| \frac{z - 8 + 2i}{z + 2i} \right| = \frac{3}{5} \right\}$. Then Σz in S |z|²$\sum_{z \in S} |z|^2$ is equal to
A.398$398$
B.413$413$
C.423$423$
D.385$385$
Solution
Related Formula
|z - z₁| = |z - z₂| represents the perpendicular bisector of the segment joining z₁ and z₂$$|z - z_1| = |z - z_2| \text{ represents the perpendicular bisector of the segment joining } z_1 \text{ and } z_2$$|x+iy|² = x² + y²$$|x+iy|^2 = x^2 + y^2$$
Core Logic
First condition: |z - 6i| = |z - 2i|$|z - 6i| = |z - 2i|$.
This means z$z$ lies on the perpendicular bisector of (0,6)$(0,6)$ and (0,2)$(0,2)$.
Let z = x + iy$z = x + iy$. Thus, y = 4$y = 4$.
Step 1: Circle Equation
Second condition: 5|z - 8 + 2i| = 3|z + 2i|$5|z - 8 + 2i| = 3|z + 2i|$.
Substitute y = 4$y = 4$ into z$z$: z = x + 4i$z = x + 4i$.
Whenever an absolute value ratio equals 1, immediately map it to a line (perpendicular bisector) and substitute its constraint directly into the second curve equation to reduce dimensionality.
Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
Q23jee_main_2026_24_january_eveningProperties of Moduli
Let z = (1 + i)(1 + 2i)(1 + 3i) (1 + ni)$z = (1 + \mathrm{i})(1 + 2\mathrm{i})(1 + 3\mathrm{i}) \dots (1 + \mathrm{ni})$, where i = √(-1)$\mathrm{i} = \sqrt{-1}$. If |z|² = 44200$|z|^2 = 44200$, then n$n$ is equal to
The calculated product for n=5$n=5$ exactly matches the prime factorization of 44200$44200$.
Therefore, n = 5$n = 5$.
Pattern Recognition
Modulus is multiplicative. In problems featuring chains of complex multiplications set equal to a huge real magnitude, instantly switch to magnitudes and map to integer factorization.
Chapter Mix
Class 11 Maths: Complex Numbers
Q5jee_main_2026_28_january_morningGeometry of Complex Numbers
Let z$z$ be a complex number such that |z - 6| = 5$|z - 6| = 5$ and |z + 2 - 6i| = 5$|z + 2 - 6i| = 5$. Then the value of z³ + 3z² - 15z + 141$z^{3} + 3z^{2} - 15z + 141$ is equal to
A.42$42$
B.37$37$
C.50$50$
D.61$61$
Solution
Core Logic
Geometry of Complex Numbers
The given equations represent two circles in the complex plane:
Circle 1: Center C₁(6, 0)$C_1(6, 0)$, radius r₁ = 5$r_1 = 5$
Circle 2: Center C₂(-2, 6)$C_2(-2, 6)$, radius r₂ = 5$r_2 = 5$
When given two complex distance modulus equations |z-z₁|=r₁$|z-z_1|=r_1$ and |z-z₂|=r₂$|z-z_2|=r_2$, always check the distance between centers |z₁ - z₂|$|z_1 - z_2|$. If it exactly equals r₁ + r₂$r_1 + r_2$, the single unique solution is the section formula midpoint.
Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
Q10jee_main_2026_28_january_morningNature of Roots
If α, β$\alpha, \beta$, where α < β$\alpha < \beta$, are the roots of the equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$ such that (1)/(α) - (1)/(β) = (1)/(3)$\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, then the sum of all possible values of λ$\lambda$ is:
A.6$6$
B.2$2$
C.4$4$
D.8$8$
Solution
Related Formula
For a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$:
Sum of roots: α + β = -(b)/(a)$\alpha + \beta = -\frac{b}{a}$
Product of roots: αβ = (c)/(a)$\alpha\beta = \frac{c}{a}$
Core Logic
From the given equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$:
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.
