Complex Numbers
De Moivre's Theorem ā Cube Roots of Unity
Complex Numbers_PYQ
Grade 11
Question:
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
$\dfrac{\pi}{4}$
$\dfrac{\pi}{6}$
$0$
$\dfrac{\pi}{3}$
Step-by-Step Solution
Key Concept: $\omega^3=1$ so exponents reduce $\pmod{3}$. Both $81$ and $93$ are divisible by $3$, giving $z_0^{81}=z_0^{93}=1$.
**Step 1: Identify zā**
$x^2+x+1=0$ has roots $\omega, \omega^2$ with $\omega^3=1$.
**Step 2: Reduce powers mod 3**
$z_0^{81} = (z_0^3)^{27} = 1$. $\quad z_0^{93} = (z_0^3)^{31} = 1$.
**Step 3: Compute z and its argument**
$z = 3 + 6i - 3i = 3 + 3i$. $\quad \arg(3+3i) = \arctan(1) = \dfrac{\pi}{4}$.
Correct Answer: 1