Sum of common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{1985}+z^{100}+1=0$ is
- $-1$
- $1$
- $0$
- $2$
The equation z^3+2z^2+2z+1=0 factors as (z+1)(z^2+z+1)=0. The roots are -1 and the complex cube roots of unity (omega, omega^2). Testing these in the second equation z^1985+z^100+1=0: for z=-1, (-1)^1985+(-1)^100+1 = -1+1+1 = 1 (not a root). For z=omega, omega^1985+omega^100+1 = omega^2+omega+1 = 0 (is a root). For z=omega^2, (omega^2)^1985+(omega^2)^100+1 = omega^4000+omega^200+1 = omega+omega^2+1 = 0 (is a root). The common roots are omega and omega^2, and their sum is -1.
Factoring the first equation z^3 + 2z^2 + 2z + 1 = 0 by grouping gives z^2(z + 1) + (z + 1)^2 = 0. This factors further into (z + 1)(z^2 + z + 1) = 0, meaning the roots are -1 and the complex cube roots of unity. We substitute these roots into the second equation, z^1985 + z^100 + 1 = 0, to find the common roots. Since 1985 and 100 are not multiples of 3, the complex roots do not satisfy the second equation; however, substituting z = -1 gives (-1)^1985 + (-1)^100 + 1 = -1 + 1 + 1 = 1, which is not zero. The only common root is found by testing z such that z^2+z+1=0, so z^3=1. For z^3=1, z^1985 = z^(3*661+2) = z^2, and z^100 = z^(3*33+1) = z. The equation becomes z^2 + z + 1 = 0, which is perfectly satisfied by the complex roots omega and omega^2. Since there are two such roots forming the pair (omega, omega^2), their sum is -1.