Multiple choice

Let ${z}{1}$ and ${z}{2}$ be roots of the equation ${z}^{2}+pz+q=0,$ where the coefficients $p$ and $q$ may be complex numbers. Let $A$ and $B$ represents ${z}{1}$ and ${z}{2}$ in the complex plane. If $\angle AOB=\alpha \neq 0$ and $OA=OB,$ where $O$ is the origin, then $\displaystyle{ p }^{ 2 }=k\cos ^{ 2 }{ \frac { \alpha }{ 2 } } ,$ where $k=$

  1. $q$
  2. $2q$
  3. $4q$

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

The roots are z1 = r*exp(i*theta1) and z2 = r*exp(i*theta2) with |theta1 - theta2| = alpha. Then p = -(z1 + z2) and q = z1*z2. Using the identity (z1 + z2)^2 = (z1 - z2)^2 + 4*z1*z2, and noting (z1 - z2)^2 = r^2 * (exp(i*theta1) - exp(i*theta2))^2 = r^2 * exp(i*(theta1+theta2)) * (exp(i*alpha/2) - exp(-i*alpha/2))^2 * exp(-i*alpha) = -4*r^2 * exp(i*(theta1+theta2)) * sin^2(alpha/2). Since q = r^2 * exp(i*(theta1+theta2)), p^2 = 4*q*sin^2(alpha/2) - 4*q*cos^2(alpha/2) is not quite right; the standard derivation leads to p^2 = 4q*cos^2(alpha/2).

AI explanation

Using Vieta's formulas, we know that z1 + z2 = -p and z1 * z2 = q. The square of the sum gives p^2 = (z1 + z2)^2 = z1^2 + z2^2 + 2z1z2 = q(z1/z2 + z2/z1 + 2) = q(z1^2 + z2^2 + 2z1z2)/(z1z2) = q(z1+z2)^2/(z1z2) = q(p^2/q) = p^2. Since OA = OB, the magnitude of z1/z2 is 1, meaning z1/z2 lies on the unit circle and can be written as e^(i alpha) = cos(alpha) + i sin(alpha). We rewrite p^2 from before as p^2 = q(z1/z2 + z2/z1 + 2) = q(cos(alpha) + i sin(alpha) + cos(alpha) - i sin(alpha) + 2) = q(2cos(alpha) + 2). By the half-angle identity, 1 + cos(alpha) = 2cos^2(alpha/2), so p^2 = 4q cos^2(alpha/2). Matching this to p^2 = k cos^2(alpha/2), k = 4q.

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