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MathsMediumClass 11

De Moivre's Theorem — nth roots, roots of unity

Complex Numbers

8

JEE Qs

8%

Hard

75

min

Thoroughly understand the geometric interpretation of De Moivre's Theorem and roots of unity; this often simplifies complex algebraic problems into visual ones.

🧮 Key Formulas

(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)
z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] where z = r(cos θ + i sin θ) and k = 0, 1, ..., n-1
nth roots of unity (z^n = 1): ω_k = cos((2kπ)/n) + i sin((2kπ)/n) for k = 0, 1, ..., n-1
Sum of nth roots of unity (n > 1): 1 + ω + ω^2 + ... + ω^(n-1) = 0
Product of nth roots of unity: 1 * ω * ω^2 * ... * ω^(n-1) = (-1)^(n-1)
If ω is an nth root of unity, then ω^n = 1

✅ Key Points for JEE

  • 1De Moivre's Theorem simplifies finding integer powers and rational roots of complex numbers when expressed in polar form.
  • 2For finding 'n' distinct nth roots, it is crucial to use the general argument (θ + 2kπ) before dividing by 'n', where k takes values from 0 to n-1.
  • 3The 'n' distinct nth roots of any complex number z = r(cos θ + i sin θ) form the vertices of a regular n-sided polygon inscribed in a circle of radius r^(1/n) centered at the origin.
  • 4The nth roots of unity are equally spaced points on the unit circle in the Argand plane, forming a regular n-gon with one vertex at (1,0).
  • 5Master the summation and product properties of roots of unity, as they are frequently used in problems involving polynomial equations and series.

⚠️ Common Mistakes

  • ✕Forgetting to include the '2kπ' term in the argument when finding nth roots, leading to only one root instead of 'n' distinct roots.
  • ✕Incorrectly identifying the range of 'k' for distinct roots; it should be k = 0, 1, ..., n-1 (or any 'n' consecutive integers).
  • ✕Errors in converting the given complex number into its correct polar form (magnitude 'r' and principal argument 'θ').

📝 Practice Questions

See all

Q17.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

2025·MCQMedium

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

2025·NumericalHard

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

2025·MCQHard

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

2025·MCQHard

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

2025·MCQMedium

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

2025·MCQHard

NCERT Chapters

  • Class 11 Maths Ch 5: Complex Numbers and Quadratic Equations