Dabour Hub EGEgyptian Curriculum 2026/2027
2-3 The cube roots of unity

2-3 The cube roots of unity

Point 8 of 8 — Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres TheoremQuestion bank: 7 questions (7 with pictures)

08 2-3 The cube roots of unity

Finding them

  • Solve z3=1 with De Moivre's theorem: the roots are 1, −12+32i and −12−32i.
  • Algebraic check: z3−1=(z−1)(z2+z+1)=0 gives z=1 or z=−1±3 i2.
  • The square of one complex root equals the other, so the roots are written as 1, ω, ω2 where: ω=−12+32i,ω2=−12−32i
  • On the Argand plane the roots lie on the unit circle, at angles 0∘, 120∘ and −120∘, and form an equilateral triangle.

Properties of the cube roots of unity

  • The sum of the roots is zero: 1+ω+ω2=0
  • Hence 1+ω=−ω2,  1+ω2=−ω,  ω+ω2=−1.
  • The product is one: ω3=1, so ω3n=1, ω3n+1=ω and ω3n+2=ω2 for any integer n.
  • Reciprocals: 1ω=ω2 and 1ω2=ω; also ωˉ=ω2.
  • The difference of the two complex roots: ω−ω2=3 i and ω2−ω=−3 i.
  • All three roots have modulus 1: ∣1∣=∣ω∣=∣ω2∣=1.
  • To simplify a power of ω, divide the exponent by 3 and keep the remainder.
✍️ Worked examples

Example 19: Find the value of 5+5ω+5ω2 and of (1−2ω−2ω2)(3+5ω+5ω2).

Solution:

  • 5+5ω+5ω2=5(1+ω+ω2)=5×0=0.
  • Use 1ω=ω2 and 1ω2=ω.
  • First bracket: 1−2ω2−2ω=1−2(ω+ω2)=1+2=3.
  • Second bracket: 3+5(ω+ω2)=3−5=−2.
  • The product is 3×(−2)=−6.

Applications

  • Forming a quadratic equation from two roots that are expressed in ω: find the sum and product of the roots using 1+ω+ω2=0 and ω3=1, then write x2−(sum)x+(product)=0.
  • Reducing expressions: replace 1/ω by ω2, group terms into 1+ω+ω2 and reduce powers modulo 3.
  • Complex numbers are used in many fields, including electricity, dynamics, relativity and other physical applications.

📝 Question bank for this point

7 questions linked to this point only, 7 of them with pictures.