Cube Roots of Unity — ω, ω² properties
Complex Numbers
8
JEE Qs
8%
Hard
75
min
Master the properties 1+ω+ω²=0 and ω^3=1, as most problems reduce to clever application of these two fundamental identities.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always simplify powers of ω by dividing the exponent by 3 and using the remainder (e.g., ω^n = ω^(n mod 3)) based on ω^3 = 1.
- 2The identity 1 + ω + ω^2 = 0 is fundamental for simplifying expressions involving sums of cube roots of unity.
- 3Recognize that the cube roots of unity (1, ω, ω²) represent the vertices of an equilateral triangle inscribed in the unit circle in the Argand plane.
- 4Familiarize yourself with the factorization identity a^3+b^3+c^3-3abc = (a+b+c)(a+bω+cω²)(a+bω²+cω) as it is frequently tested.
- 5Expressions like (1+ω)^n can be simplified as (-ω²)^n and (1+ω²)^n as (-ω)^n, making calculations much faster.
⚠️ Common Mistakes
- ✕Confusing the exact values of ω and ω² (e.g., misplacing the positive or negative sign for i√3).
- ✕Errors in simplifying ω^n by not correctly applying the modulo 3 rule for the exponent.
- ✕Forgetting to use 1 + ω + ω² = 0 when simplifying sums, or misapplying it to terms not involving ω or ω².
- ✕Applying the property 'a + bω + cω² = 0 implies a = b = c' without ensuring a, b, c are real numbers.
📝 Practice Questions
See allQ17.Let 2¯z+i ¯z−i = 13 , z ∈C , be the equation of a circle with center at C . If the area of the triangle, whose vertices are at the points (0, 0), C and (α, 0) is 11 square units, then α2 equals: (1) 50 (2) 100 (3) 81 (4) 121 25 25
Q25.Let α, β be the roots of the equation x2 −ax −b = 0 with Im(α) < Im(β). Let Pn = αn −βn . If P3 = −5√7i, P4 = −3√7i, P5 = 11√7i and P6 = 45√7i , then α4 + β4 is equal to . ∣∣ 2025 (23 Jan Shift 2) JEE Main Previous Year Paper
Q20.Let z1, z2 and z3 be three complex numbers on the circle |z| = 1 with arg (z1) = −π4 , arg (z2) = 0 and arg (z3) = π4 . If |z1¯z2 + z2¯z3 + z3¯z1|2 = α + β√2, α, β ∈Z, then the value of α2 + β2 is : (1) 24 (2) 29 (3) 41 (4) 31
Q19.Let the curve z(1 + i) + ¯z(1 −i) = 4, z ∈C, divide the region |z −3| ≤1 into two parts of areas α and β . Then |α −β| equals : (1) 1 + π2 (2) 1 + π3 (3) 1 + π6 (4) 1 + π4
Q14.The number of complex numbers z , satisfying |z| = 1 and z¯z + ¯zz = 1, is : (1) 4 (2) 8 (3) 10 (4) 6 Q15. ⎡ 0 ⎤ ⎡ 0 ⎤ ⎡4⎤ ⎡0⎤ ⎡2 ⎤ ⎡1 ⎤ Let A = [aij] be 3 × 3 matrix such that A 1 = 0 , A 1 = 1 and A 1 = 0 , then a23 equals : ⎣ 0 ⎦ ⎣ 1 ⎦ ⎣3⎦ ⎣0⎦ ⎣2 ⎦ ⎣0 ⎦ (1) -1 (2) 2 (3) 1 (4) 0 2 x sin 2 dx equals : 3 3
Q10.For a statistical data x1, x2, … , x10 of 10 values, a student obtained the mean as 5.5 and ∑10i=1 x2i = 371. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is (1) 9 (2) 5 (3) 7 (4) 4
NCERT Chapters
- Class 11 Maths Ch 5: Complex Numbers and Quadratic Equations