Multiple choice

Let $z_0$ be a root of the quadratic equation $x^2+x+1=0$. If $z=3+6iz_0^{81}-3iz_0^{93}$, then arg $z$ is equal to :

  1. $\dfrac{\pi}{4}$
  2. $\dfrac{\pi}{3}$
  3. $0$
  4. $\dfrac{\pi}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

z0 is a root of x^2 + x + 1 = 0, so z0 = omega or omega^2. omega^3 = 1. z0^81 = (omega^3)^27 = 1. z0^93 = (omega^3)^31 = 1. z = 3 + 6i(1) - 3i(1) = 3 + 3i. arg(z) = arctan(3/3) = pi/4.

AI explanation

Since z0 is a root of x^2 + x + 1 = 0, it is a complex cube root of unity, meaning z0^3 = 1. This allows us to simplify the powers z0^81 and z0^93 to 1, reducing the expression for z to 3 + 6i(1) - 3i(1), which simplifies to z = 3 + 3i. The complex number z lies in the first quadrant with real and imaginary parts of 3, so its argument is calculated as arctan(3 / 3) = pi/4.

Keep practicing — related questions