Multiple choice

Let $x_1, x_2,$ are the roots of quadratic equation $x^2 + ax + b = 0$, Where $ a, b$ are complex numbers and $y_1, y_2$ are the roots of the quadratic equation $y^2 + \left|a\right| y + \left|b\right| =0$. If $\left|x_1\right| = \left|x_2\right| = 1$, then

  1. $\left|y_1\right| = \left|y_2\right|= 1$.
  2. $\left|y_1\right| = \left|y_2\right|\neq 1$.
  3. $\left|y_1\right| \neq 1, \left|y_2\right|= 1$.
  4. $\left|y_1\right| =1, \left|y_2\right|\neq 1$.
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A Correct answer
Explanation

Given |x1| = |x2| = 1, the product of roots |x1*x2| = |b| = 1. The equation for y is y^2 + |a|y + 1 = 0. Since the roots of y^2 + |a|y + 1 = 0 have a product of 1, if one root has magnitude 1, the other must also have magnitude 1 (since |y1*y2| = 1).

AI explanation

Given that the roots x1 and x2 both have an absolute value of 1, the magnitude of their product is 1, so |b| = |x1 * x2| = 1. Additionally, the magnitude of their sum equals the absolute value of negative a, so |a| = |x1 + x2| <= |x1| + |x2| = 2. We substitute these values into the second equation to get y^2 + |a|y + 1 = 0. The roots y1 and y2 must satisfy |y1 * y2| = |1| = 1 from Vieta's formulas. Let y1 = r(cos(a) + i sin(a)), making the conjugate y2 = (1/r)(cos(a) - i sin(a)); the sum |a| = (r + 1/r)cos(a) must be real, and for y^2 + cy + 1 = 0 to have roots with a product of 1, their magnitudes must be 1.

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