Multiple choice

If ${z}{1}$ and ${z}{2}$ are two complex roots of equal magnitude and their arguments differ by $\displaystyle \frac { \pi }{ 2 } $, of the quadratic equation $a{ x }^{ 2 }+bx+c=0\left( a\neq 0 \right) $ then $a($ in terms of $b$ and $c)$ is

  1. $\displaystyle \frac { { b }^{ 2 } }{ 2c } $
  2. $\displaystyle \frac { { b }^{ 2 } }{ c } $
  3. $\displaystyle \frac { { b } }{ 2c } $

  4. None of these
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A Correct answer
Explanation

Let roots be r*exp(i*theta) and r*exp(i*(theta + pi/2)). Product of roots c/a = r^2 * exp(i*(2*theta + pi/2)) = r^2 * i * exp(i*2*theta). Sum of roots -b/a = r*exp(i*theta) * (1 + i). Squaring: b^2/a^2 = r^2 * exp(i*2*theta) * (1 + i)^2 = r^2 * exp(i*2*theta) * 2i. Thus, b^2/a^2 = 2 * (c/a) = 2c/a. a = b^2 / 2c.

AI explanation

Let the roots be z1 = r(cos t + i sin t) and z2 = r(cos(t + pi/2) + i sin(t + pi/2)). The sum of the roots gives b/a = z1 + z2 = r(cos t - sin t) + i r(cos t + sin t), and the product gives c/a = z1 z2 = r^2 i. Squaring the sum results in (b/a)^2 = r^2(cos t - sin t)^2 - r^2 + 2 i r^2(cos t - sin t)(cos t + sin t). Since the real part of z1 + z2 squared equals the imaginary part, this confirms b^2 / a^2 = 2 i r^2. Substituting i r^2 = c/a yields b^2 / a^2 = 2c / a, which simplifies to a = b^2 / (2c).

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