Solution
Related Formula
z³-1 = (z-1)(z²+z+1) Cube roots of unity: 1, ω, ω² ω³ = 1 and 1+ω+ω² = 0Core Logic
Let's first find the roots of the lower degree polynomial:
z³ + 2z² + 2z + 1 = 0Group the terms:
(z³ + 1) + 2z(z + 1) = 0 (z + 1)(z² - z + 1) + 2z(z + 1) = 0 (z + 1)(z² - z + 1 + 2z) = 0 (z + 1)(z² + z + 1) = 0Thus, the roots are z = -1, and the roots of z² + z + 1 = 0 which are z = ω, ω².
Step 1: Check z = -1
Substitute z = -1 into the first equation z¹⁹⁸⁵ + z¹⁰⁰ + 1 = 0:
(-1)¹⁹⁸⁵ + (-1)¹⁰⁰ + 1 = -1 + 1 + 1 = 1 ≠ 0So, z = -1 is not a common root.
Step 2: Check z = ω and ω²
Substitute z = ω:
ω¹⁹⁸⁵ + ω¹⁰⁰ + 1Reduce the powers modulo 3 (since ω³ = 1): 1985 = 3 × 661 + 2 ⇒ ω¹⁹⁸⁵ = ω² 100 = 3 × 33 + 1 ⇒ ω¹⁰⁰ = ω¹ = ω So, ω¹⁹⁸⁵ + ω¹⁰⁰ + 1 = ω² + ω + 1 = 0. z = ω is a common root.
Substitute z = ω²:
(ω²)¹⁹⁸⁵ + (ω²)¹⁰⁰ + 1 = ω³⁹⁷⁰ + ω²⁰⁰ + 13970 = 3 × 1323 + 1 ⇒ ω³⁹⁷⁰ = ω 200 = 3 × 66 + 2 ⇒ ω²⁰⁰ = ω² So, ω³⁹⁷⁰ + ω²⁰⁰ + 1 = ω + ω² + 1 = 0. z = ω² is also a common root.
Step 3: Conclusion
There are exactly 2 common roots: ω and ω².
Pattern Recognition
Whenever z²+z+1 emerges as a factor, its roots ω and ω² can be directly tested in massive degree polynomials using exponent reduction modulo 3.
Chapter Mix
Class 11 Maths: Complex Numbers