Question

Let z₁ and z₂ be any two distinct complex number which satisfy $z_{1}^{20} = 1$.

3. The probability that $|z₁ - z₂| \ge \frac{\sqrt{5} + 1}{2}$ is $\frac{p}{q}$ (where p and q are co-prime) then the value of q - 2p is :

4. The number of possible ordered pair $(z₁, z₂)$ such that $|z₁ - z₂| \le \sqrt{2 - \sqrt{3}}$

Answer 1

1

Answer 2

Let $z$ be a complex number satisfying $z^{20} = 1$. The solutions are the 20th roots of unity, given by $z_k = e^{i \frac{2\pi k}{20}} = e^{i \frac{\pi k}{10}}$ for $k = 0, 1, \dots, 19$. These roots are points on the unit circle in the complex plane.

Problem 3: Probability that $|z_1 - z_2| \ge \frac{\sqrt{5} + 1}{2}$.

$z_1$ and $z_2$ are distinct roots of $z^{20}=1$. The total number of distinct unordered pairs $\{z_1, z_2\}$ is $\binom{20}{2} = \frac{20 \times 19}{2} = 190$. This is our sample space size.

Let $z_j = e^{i \frac{2\pi j}{20}}$ and $z_k = e^{i \frac{2\pi k}{20}}$ be two distinct roots, $j, k \in \{0, 1, \dots, 19\}$, $j \ne k$. The distance between $z_j$ and $z_k$ is given by $|z_j - z_k|$. $|z_j - z_k| = |e^{i \frac{2\pi j}{20}} - e^{i \frac{2\pi k}{20}}| = |e^{i \frac{\pi (j+k)}{20}} (e^{i \frac{\pi (j-k)}{20}} - e^{-i \frac{\pi (j-k)}{20}})| = |e^{i \frac{\pi (j+k)}{20}}| |2i \sin(\frac{\pi (j-k)}{20})| = 2 |\sin(\frac{\pi (j-k)}{20})|$. Since the roots are on the unit circle, the distance between two roots depends only on the angular separation between them. The angular separation is $\frac{2\pi (k-j)}{20}$. The distance is $2 \sin(\frac{\pi |k-j|}{20})$. Due to symmetry on the circle, the distance depends on the shorter arc between the two points. Let $d = |k-j|$. The relevant difference is $m = \min(d \pmod{20}, 20 - (d \pmod{20}))$. $m$ represents the number of steps (of size $2\pi/20$) along the shorter arc on the circle. $m$ can take values $1, 2, \dots, 10$. The distance is $2 \sin(\frac{\pi m}{20})$.

We need to find the pairs $\{z_1, z_2\}$ such that $2 \sin(\frac{\pi m}{20}) \ge \frac{\sqrt{5} + 1}{2}$. This is equivalent to $\sin(\frac{\pi m}{20}) \ge \frac{\sqrt{5} + 1}{4}$. We know that $\cos(\frac{\pi}{5}) = \cos(36^\circ) = \frac{\sqrt{5}+1}{4}$. Also, $\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$. So $\sin(\frac{3\pi}{10}) = \sin(\frac{\pi}{2} - \frac{\pi}{5}) = \frac{\sqrt{5}+1}{4}$. $\frac{3\pi}{10} = \frac{6\pi}{20}$. So the condition is $\sin(\frac{\pi m}{20}) \ge \sin(\frac{6\pi}{20})$. Since $m \in \{1, 2, \dots, 10\}$, the angle $\frac{\pi m}{20}$ is in the range $(0, \pi/2]$. In this range, the sine function is increasing. Thus, the inequality is equivalent to $\frac{\pi m}{20} \ge \frac{6\pi}{20}$, which means $m \ge 6$. The possible values for $m$ are $6, 7, 8, 9, 10$.

For a given value of $m \in \{1, 2, \dots, 9\}$, there are 20 unordered pairs $\{z_j, z_k\}$ such that the minimum index difference is $m$. These are pairs $\{z_j, z_{j+m}\}$ for $j=0, \dots, 19$ (indices mod 20). For $m \in \{1, \dots, 9\}$, $m \ne 20-m$, so $\{z_j, z_{j+m}\}$ and $\{z_j, z_{j-m}\}$ represent the same set of 20 pairs due to symmetry. For $m=10$, $m = 20-m$. The pairs $\{z_j, z_{j+10}\}$ are diametrically opposite points. There are 10 such unordered pairs: $\{z_0, z_{10}\}, \{z_1, z_{11}\}, \dots, \{z_9, z_{19}\}$.

The values of $m$ satisfying $m \ge 6$ are $m=6, 7, 8, 9, 10$. Number of favourable pairs: For $m=6$: 20 pairs. Distance $2 \sin(6\pi/20) = (\sqrt{5}+1)/2$. For $m=7$: 20 pairs. Distance $2 \sin(7\pi/20) > (\sqrt{5}+1)/2$. For $m=8$: 20 pairs. Distance $2 \sin(8\pi/20) > (\sqrt{5}+1)/2$. For $m=9$: 20 pairs. Distance $2 \sin(9\pi/20) > (\sqrt{5}+1)/2$. For $m=10$: 10 pairs. Distance $2 \sin(10\pi/20) = 2$.

Total number of favourable pairs = $20 + 20 + 20 + 20 + 10 = 90$. Total number of distinct unordered pairs = 190. The probability is $\frac{90}{190} = \frac{9}{19}$. This is given as $\frac{p}{q}$ where $p$ and $q$ are co-prime. $p=9$, $q=19$. They are co-prime. We need to find the value of $q - 2p$. $q - 2p = 19 - 2(9) = 19 - 18 = 1$.

Problem 4: The number of possible ordered pairs $(z_1, z_2)$ such that $|z_1 - z_2| \le \sqrt{2 - \sqrt{3}}$.

$z_1$ and $z_2$ are distinct roots of $z^{20}=1$. The total number of distinct ordered pairs $(z_1, z_2)$ is $20 \times 19 = 380$.

The condition is $|z_1 - z_2| \le \sqrt{2 - \sqrt{3}}$. We know $|z_1 - z_2| = 2 \sin(\frac{\pi m}{20})$, where $m = \min(|k-j| \pmod{20}, 20-|k-j| \pmod{20})$ for $z_1=z_j$ and $z_2=z_k$. $m \in \{1, 2, \dots, 10\}$. We need $2 \sin(\frac{\pi m}{20}) \le \sqrt{2 - \sqrt{3}}$. Let's evaluate $\sqrt{2 - \sqrt{3}}$. $\sqrt{2 - \sqrt{3}} = \sqrt{\frac{4 - 2\sqrt{3}}{2}} = \frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}} = \frac{\sqrt{3}-1}{\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{2}$. We know $\sin(\frac{\pi}{12}) = \sin(15^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4}$. So $\sqrt{2 - \sqrt{3}} = 2 \sin(\frac{\pi}{12})$. $\frac{\pi}{12} = \frac{20\pi}{240} = \frac{\pi \times (20/12)}{20} = \frac{\pi \times 5/3}{20}$. This is not directly in the form $\frac{\pi m}{20}$. Let's express $\frac{\pi}{12}$ in terms of $\frac{\pi}{20}$. $\frac{\pi}{12} = \frac{5\pi}{60} = \frac{5\pi \times 3}{180} = \frac{5\pi}{12}$. $\frac{\pi}{12} = \frac{20\pi/12}{20} = \frac{5\pi/3}{20}$. The condition is $2 \sin(\frac{\pi m}{20}) \le 2 \sin(\frac{\pi}{12})$. $\sin(\frac{\pi m}{20}) \le \sin(\frac{\pi}{12})$. $\frac{\pi}{12} = 15^\circ$. The angles $\frac{\pi m}{20}$ for $m=1, \dots, 10$ are $\frac{\pi}{20}=9^\circ, \frac{2\pi}{20}=18^\circ, \dots, \frac{10\pi}{20}=90^\circ$. Since $\frac{\pi m}{20}$ is in $(0, \pi/2]$ for $m \in \{1, \dots, 10\}$, the inequality $\sin(\frac{\pi m}{20}) \le \sin(\frac{\pi}{12})$ is equivalent to $\frac{\pi m}{20} \le \frac{\pi}{12}$. $m/20 \le 1/12 \implies m \le 20/12 = 5/3$. Since $m$ must be an integer, the possible values for $m$ are $m=1$. The condition $|z_1 - z_2| \le \sqrt{2 - \sqrt{3}}$ is satisfied if and only if $m=1$. The minimum index difference between $z_1$ and $z_2$ must be 1. This means $z_1$ and $z_2$ must be adjacent roots on the unit circle. The pairs of adjacent roots are $(z_0, z_1), (z_1, z_2), \dots, (z_{19}, z_0)$. There are 20 such ordered pairs. Also, $(z_1, z_0), (z_2, z_1), \dots, (z_0, z_{19})$ are ordered pairs of adjacent roots. There are 20 such ordered pairs. The minimum index difference $m=1$ corresponds to pairs $(z_j, z_{j+1})$ or $(z_j, z_{j-1})$ (indices mod 20). The ordered pairs $(z_1, z_2)$ with minimum index difference 1 are $(z_j, z_{j+1})$ for $j=0, \dots, 19$ and $(z_{j+1}, z_j)$ for $j=0, \dots, 19$. There are 20 pairs of the form $(z_j, z_{j+1})$ and 20 pairs of the form $(z_{j+1}, z_j)$. Total number of ordered pairs $(z_1, z_2)$ such that $|z_1 - z_2| = 2 \sin(\pi/20)$ (which is the distance for $m=1$) is $20+20=40$. Since $m=1$ is the only value that satisfies the condition, the number of possible ordered pairs $(z_1, z_2)$ is 40.

The final answer is $\boxed{1}$.