Multiple choice

The common roots of the equation ${ z }^{ 3 }+2{ z }^{ 2 }+2z+1=0$ and ${ z }^{ 1985 }+{ z }^{ 100 }+1=0$ are

  1. $-1,\omega $
  2. $-1,{ \omega  }^{ 2 }$
  3. $\omega ,{ \omega  }^{ 2 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation z^3 + 2z^2 + 2z + 1 = 0 can be factored as (z+1)(z^2+z+1) = 0. The roots are -1, omega, and omega^2. Testing these in z^1985 + z^100 + 1 = 0: for omega, omega^1985 + omega^100 + 1 = omega^2 + omega + 1 = 0. Thus, omega and omega^2 are common roots.

AI explanation

The first equation factors as (z+1)(z^2+z+1)=0, giving roots of -1, omega, and omega^2. For the second equation, z must be either omega or omega^2, because replacing -1 yields (-1)^1985 + (-1)^100 + 1, which equals 1 instead of 0. Testing the complex cube roots of unity in the second equation gives omega^1985 + omega^100 + 1 = omega^2 + omega + 1 = 0, and similarly for omega^2. Therefore, the common roots are omega and omega^2.

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