Multiple choice

If $z_1,z_2,z_3,$ and $z_4$ are the complex roots of the equation $z^4+3z^2+1=0$, then the value of $\prod { i=1 }^{ 4 }{ 4+\left( z{ i }^{ 2 } \right) } $ is

  1. $9$
  2. $12$
  3. $25$
  4. $27$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of z^4 + 3z^2 + 1 = 0 satisfy z^2 = (-3 +/- sqrt(9-4))/2 = (-3 +/- sqrt(5))/2. Let the roots be a, b, c, d. The product is (4+a^2)(4+b^2)(4+c^2)(4+d^2). Using the polynomial P(z) = z^4 + 3z^2 + 1, we evaluate P(2i) or similar substitutions.

AI explanation

The expression to evaluate is the product of (4 + zi squared) for all four roots of the polynomial z to the fourth + 3z squared + 1 = 0. If we define a new variable u = z squared, the given equation becomes u squared + 3u + 1 = 0, which has two roots u1 and u2. For each root ui of this quadratic, the original polynomial has two roots of z, so zi squared takes the values u1 and u2, each exactly twice. The desired product is therefore (4 + u1) squared (4 + u2) squared. We can rewrite this as the square of the product (4 + u1)(4 + u2). Using the identity (x + u1)(x + u2) = x squared - (sum of roots)x + (product of roots), we substitute x = 4. The sum of the roots is -3 and the product is 1, so (4 + u1)(4 + u2) = 16 - 12 + 1 = 5. Squaring this result gives 25.

Keep practicing — related questions