If z is a complex number, then the number of common roots of the equations z¹⁹⁸⁵+z¹⁰⁰+1=0 and z³+2 z²+2 z+1=0, is equal to

Solution

Maths sir · 3 steps

Factor the cubic first, then test each of its three roots in the big equation using powers of ω.

1. Factor the cubic. Group the terms: z³ + 1 + 2z(z+1) = (z+1)(z² - z + 1) + 2z(z+1) = (z+1)(z² + z + 1). So its roots are -1, ω and ω².
2. Reduce the powers. Since ω³ = 1, only the remainder of each exponent on division by 3 matters. 1985 = 3 × 661 + 2 and 100 = 3 × 33 + 1.

   Root	z¹⁹⁸⁵	z¹⁰⁰	z¹⁹⁸⁵+z¹⁰⁰+1
   -1	-1	1	1 ≠ 0
   ω	ω²	ω	ω²+ω+1 = 0
   ω²	ω³⁹⁷⁰ = ω	ω²⁰⁰ = ω²	ω+ω²+1 = 0

3. So. Only ω and ω² satisfy both equations, so there are 2 common roots, option (b).

The idea to keep. When you see z²+z+1 hiding in a polynomial, think ω and ω². Cut every exponent down to its remainder mod 3, and always test both roots and any real root separately.

Answer from the NTA final answer key. Solution by PrepWiser.