Question

$z_1, z_2,........z_9$ are roots of the equation $z^9 + \alpha z^8 + \alpha^2z^7 +........+ \alpha^8z + \alpha^9 = 0$ ($z \in C$, $\alpha \in R^+$), then

• A. $\sum_{r=0}^{37}\sum_{i=1}^{9} z_i^r = 2$ for $\alpha = 1$
• B. $\beta_1, \beta_2 \& \beta_3 \in \{z_1, z_2, ..... z_9\}$ are vertices of a triangle such that $Arg \left(\frac{\beta_i}{\beta_j}\right) \neq \pm \frac{\pi}{5} \forall i, j \in \{1,2,3\}$ number of such triangles are 30.

Answer 1

Answer: (A)

(A) $\sum_{r=0}^{37}\sum_{i=1}^{9} z_i^r = 2$ for $\alpha = 1$

Answer 2

The roots of the equation $z^9 + \alpha z^8 + \alpha^2z^7 + \dots + \alpha^8z + \alpha^9 = 0$ can be found by multiplying the equation by $(z-\alpha)$. This gives $(z-\alpha)(z^9 + \alpha z^8 + \dots + \alpha^9) = z^{10} - \alpha^{10}$. So the equation is equivalent to $\frac{z^{10} - \alpha^{10}}{z-\alpha} = 0$. The roots are the solutions to $z^{10} - \alpha^{10} = 0$ except for $z=\alpha$. The solutions to $z^{10} = \alpha^{10}$ are $z = \alpha e^{i \frac{2k\pi}{10}} = \alpha e^{i \frac{k\pi}{5}}$ for $k = 0, 1, \dots, 9$. The root $z=\alpha$ corresponds to $k=0$. Thus, the roots of the given equation are $z_k = \alpha e^{i \frac{k\pi}{5}}$ for $k = 1, 2, \dots, 9$.

Option (A): $\sum_{r=0}^{37}\sum_{i=1}^{9} z_i^r = 2$ for $\alpha = 1$.

If $\alpha = 1$, the roots are $z_k = e^{i \frac{k\pi}{5}}$ for $k = 1, 2, \dots, 9$. Let $\zeta = e^{i \pi/5}$. The roots are $\zeta^1, \zeta^2, \dots, \zeta^9$. For any root $z_i$, we have $z_i^{10} = (e^{i k\pi/5})^{10} = e^{i 2k\pi} = 1$ for $k \in \{1, \dots, 9\}$. The inner sum is $\sum_{r=0}^{37} z_i^r$. This is a geometric series with first term 1, common ratio $z_i$, and 38 terms. Since $z_i \neq 1$ (as $k \in \{1, \dots, 9\}$), the sum is $\frac{1 - z_i^{38}}{1 - z_i}$. Since $z_i^{10} = 1$, $z_i^{38} = z_i^{3 \times 10 + 8} = (z_i^{10})^3 z_i^8 = 1^3 z_i^8 = z_i^8$. So, $\sum_{r=0}^{37} z_i^r = \frac{1 - z_i^8}{1 - z_i} = 1 + z_i + z_i^2 + \dots + z_i^7$. We need to compute $\sum_{i=1}^{9} (1 + z_i + \dots + z_i^7)$. Let $z_i = \zeta^{k_i}$ where $k_i$ are distinct values from $\{1, 2, \dots, 9\}$. The sum is $\sum_{k=1}^{9} (1 + \zeta^k + \zeta^{2k} + \dots + \zeta^{7k})$. This can be written as $\sum_{k=1}^{9} 1 + \sum_{k=1}^{9} \zeta^k + \sum_{k=1}^{9} \zeta^{2k} + \dots + \sum_{k=1}^{9} \zeta^{7k}$. The first term is $\sum_{k=1}^{9} 1 = 9$. For $m \in \{1, 2, \dots, 7\}$, consider the sum $\sum_{k=1}^{9} \zeta^{mk} = \sum_{k=1}^{9} (\zeta^m)^k$. Since $m \in \{1, 2, \dots, 7\}$, $\zeta^m = e^{i m\pi/5} \neq 1$ because $m/5$ is not an integer multiple of 2. The sum of the 10th roots of unity is $1 + \zeta^m + \zeta^{2m} + \dots + \zeta^{9m} = \sum_{k=0}^{9} (\zeta^m)^k = \frac{1 - (\zeta^m)^{10}}{1 - \zeta^m} = \frac{1 - (\zeta^{10})^m}{1 - \zeta^m} = \frac{1 - 1^m}{1 - \zeta^m} = 0$. So, $\sum_{k=1}^{9} (\zeta^m)^k = (\sum_{k=0}^{9} (\zeta^m)^k) - (\zeta^m)^0 = 0 - 1 = -1$. This holds for $m = 1, 2, \dots, 7$. The total sum is $9 + \sum_{m=1}^{7} (-1) = 9 + 7 \times (-1) = 9 - 7 = 2$. So, option (A) is correct.

Option (B): $\beta_1, \beta_2, \beta_3 \in \{z_1, z_2, \dots, z_9\}$ are vertices of a triangle such that $Arg \left(\frac{\beta_i}{\beta_j}\right) \neq \pm \frac{\pi}{5}$ for all $i, j \in \{1,2,3\}$ with $i \neq j$. Number of such triangles are 30.

Let $\beta_i = z_{k_i} = \alpha e^{i k_i\pi/5}$ and $\beta_j = z_{k_j} = \alpha e^{i k_j\pi/5}$, where $k_i, k_j \in \{1, 2, \dots, 9\}$ are distinct. $Arg \left(\frac{\beta_i}{\beta_j}\right) = Arg \left(\frac{\alpha e^{i k_i\pi/5}}{\alpha e^{i k_j\pi/5}}\right) = Arg \left(e^{i (k_i-k_j)\pi/5}\right) = (k_i-k_j)\frac{\pi}{5} \pmod{2\pi}$. The condition is $(k_i-k_j)\frac{\pi}{5} \pmod{2\pi} \neq \pm \frac{\pi}{5}$. This means $(k_i-k_j)/5 \pmod{2} \neq \pm 1/5$. $(k_i-k_j)/5 \neq \pm 1/5 + 2n$ for any integer $n$. $k_i-k_j \neq \pm 1 + 10n$. Since $k_i, k_j \in \{1, 2, \dots, 9\}$ are distinct, $k_i - k_j$ is an integer from $-8$ to $8$, excluding 0. The condition $k_i - k_j \neq \pm 1 + 10n$ for $n \in Z$. If $n=0$, the condition is $k_i - k_j \neq \pm 1$. If $n \neq 0$, $|10n| \ge 10$. Since $|k_i - k_j| \le 8$, $k_i - k_j$ cannot be equal to $\pm 1 + 10n$ for $n \neq 0$. So the condition simplifies to $|k_i - k_j| \neq 1$ for any distinct pair $i, j \in \{1, 2, 3\}$. This means that the indices $k_1, k_2, k_3$ chosen from $\{1, 2, \dots, 9\}$ must be such that no two indices are consecutive integers.

We need to choose 3 distinct indices $k_1 < k_2 < k_3$ from the set $S = \{1, 2, \dots, 9\}$ such that $k_2 - k_1 > 1$ and $k_3 - k_2 > 1$. Let $x_1 = k_1$, $x_2 = k_2 - k_1$, $x_3 = k_3 - k_2$, $x_4 = 9 - k_3$. We have $k_1 \ge 1$, $k_2 > k_1+1$, $k_3 > k_2+1$. $x_1 \ge 1$. $x_2 = k_2 - k_1 \ge 2$. $x_3 = k_3 - k_2 \ge 2$. $x_4 = 9 - k_3 \ge 9 - (9) = 0$. (It's possible that $k_3=9$) $x_1 + x_2 + x_3 + x_4 = k_1 + (k_2 - k_1) + (k_3 - k_2) + (9 - k_3) = 9$. Let $y_1 = x_1 - 1 \ge 0$, $y_2 = x_2 - 2 \ge 0$, $y_3 = x_3 - 2 \ge 0$, $y_4 = x_4 \ge 0$. $(y_1+1) + (y_2+2) + (y_3+2) + y_4 = 9$. $y_1 + y_2 + y_3 + y_4 = 9 - 1 - 2 - 2 = 4$. We need to find the number of non-negative integer solutions to $y_1 + y_2 + y_3 + y_4 = 4$. This is a stars and bars problem. The number of solutions is $\binom{n+k-1}{k-1} = \binom{4+4-1}{4-1} = \binom{7}{3}$. $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

This is the number of ways to choose 3 indices $k_1 < k_2 < k_3$ from $\{1, 2, \dots, 9\}$ such that no two are consecutive. Each set of 3 indices corresponds to a unique triangle. The number of such triangles is 35. Option (B) states the number of such triangles is 30, which is incorrect.