Multiple choice

If $z_1, z_2$ are roots of the equation $az^2+bz+c=0$, with $a, b, c > 0;\ 2b^2 > 4ac > b^2;\ z_1 \in$ Third quadrant; $z_2\in$ Second quadrant in the argand's plane then, find $arg \left (\dfrac {z_1}{z_2}\right )=$

  1. $\cos^{-1}\left (\dfrac {b^2}{2ac}\right )^{\frac12}$
  2. $2\cos^{-1}\left (\dfrac {b^2}{4ac}\right )^{\frac12}$
  3. $\cos^{-1}\left (\dfrac {b^2}{4ac}\right )^{\frac12}$
  4. $2\cos^{-1}\left (\dfrac {b^2}{2ac}\right )^{\frac12}$
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B Correct answer
Explanation

Given z1 in 3rd quadrant and z2 in 2nd quadrant, the argument of z1/z2 is arg(z1) - arg(z2). Using the properties of roots of quadratic equations and the given condition, the result simplifies to 2*cos^-1(sqrt(b^2/4ac)).

AI explanation

Let z1 = r(cos A + i sin A) in the third quadrant and z2 = r(cos B + i sin B) in the second quadrant, where both angles are measured from the positive real axis. The argument of their ratio is arg(z1 / z2) = arg(z1) - arg(z2) = A - B. From the given equation az^2 + bz + c = 0, we use the quadratic formula to find z = (-b + i sqrt(4ac - b^2)) / 2a and z = (-b - i sqrt(4ac - b^2)) / 2a. The real parts are equal to -b / 2a, so the cosine of the angle between the vectors is (-b / 2a)^2 / (c / a) = b^2 / 4ac. Since the required angle is the difference between the two arguments, the argument of z1 / z2 equals 2cos^-1(b^2 / 4ac)^1/2.

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