Probability
Roots of unity — probability of modulus condition
MJAT_TS3_P1
Grade 12
Question:
Let $z_1$ and $z_2$ be two distinct roots of the equation $z^{101}=1$. Then the probability that $|z_1+z_2|^2 \geq 2+\sqrt{3}$ is:
Step-by-Step Solution
Key Concept: $z_k=e^{2\pi i k/101}$. $|z_1+z_2|^2=2+2\cos(2\pi(m-k)/101)$. Condition: $2+2\cos\theta\geq 2+\sqrt{3}$ where $\theta=2\pi(m-k)/101$, giving $\cos\theta\geq\sqrt{3}/2\Rightarrow|\theta|\leq\pi/6\Rightarrow|m-k|\leq 101/12\approx 8.4\Rightarrow|m-k|\leq 8$.
$P=\frac{101\times 16}{101\times 100}=\frac{16}{100}=\mathbf{0.16}$.
Correct Answer: 0.16