Multiple choice

If 'a' is a complex number such that $|a| = 1$. Find the value of a, so that equation $az^2 + z + 1= 0$ has one purely imaginary root.

  1. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{5} - 1}{2} \right )$
  2. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{5} + 1}{2} \right )$
  3. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{5}}{2} \right )$
  4. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{3}}{2} \right )$
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AI explanation

Let the purely imaginary root be z = iy, where y is a non-zero real number, and let a = cos(alpha) + i sin(alpha) since |a| = 1. Substituting z into the equation gives a(iy)^2 + iy + 1 = 0, which simplifies to -y^2 cos(alpha) + 1 + i(y - y^2 sin(alpha)) = 0. Equating the imaginary part to zero gives y = 1 / sin(alpha), and equating the real part to zero gives y^2 cos(alpha) = 1. Substituting the value of y into the real part equation yields cos^2(alpha) / sin^2(alpha) = 1 / cos(alpha), which rearranges to cos^2(alpha) + cos(alpha) - 1 = 0. Solving this quadratic equation for cos(alpha) results in alpha = cos^-1( (sqrt(5) - 1) / 2 ).

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