Question

Let $\bar{z_0},\bar{z_1},\bar{z_3},...,\bar{z_n}$ be complex numbers such that $(k+1)\bar{z}_{k+1}-i(n-k)\bar{z_k}=0$ $k\epsilon\{0,1,2,...,n-1\}$. Here $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ below:

• A. $P$ is true and $Q$ is false
• B. $P$ is true and $Q$ is true
• C. $P$ is false and $Q$ is true
• D. $P$ is false and $Q$ is false

Answer 1

Answer: (C)

(C)

Answer 2

The recurrence relation is given by $(k+1)\bar{z}_{k+1}-i(n-k)\bar{z_k}=0$ for $k \in \{0, 1, ..., n-1\}$. For $n=10$, we have $(k+1)\bar{z}_{k+1}-i(10-k)\bar{z_k}=0$ for $k \in \{0, 1, ..., 9\}$. Taking the conjugate of this equation, we get $(k+1)z_{k+1}-\overline{i(10-k)\bar{z_k}}=0$. $(k+1)z_{k+1}-(-i)(10-k)z_k=0$. $(k+1)z_{k+1}+i(10-k)z_k=0$. $z_{k+1} = \frac{-i(10-k)}{k+1}z_k$ for $k=0, 1, ..., 9$.

We can express $z_k$ in terms of $z_0$ by iterating this recurrence: $z_1 = \frac{-i(10-0)}{1}z_0 = -10iz_0 = \binom{10}{1}(-i)^1 z_0$. $z_2 = \frac{-i(10-1)}{2}z_1 = \frac{-9i}{2}(-10iz_0) = \frac{90i^2}{2}z_0 = -45z_0 = \binom{10}{2}(-i)^2 z_0$. $z_3 = \frac{-i(10-2)}{3}z_2 = \frac{-8i}{3}(-45z_0) = 120iz_0 = \binom{10}{3}(-i)^3 z_0$. In general, we can show by induction that $z_k = \binom{10}{k}(-i)^k z_0$ for $k=0, 1, ..., 10$. For $k=0$, $z_0 = \binom{10}{0}(-i)^0 z_0 = 1 \cdot 1 \cdot z_0 = z_0$. Assuming $z_k = \binom{10}{k}(-i)^k z_0$ holds for some $k \in \{0, 1, ..., 9\}$, $z_{k+1} = \frac{-i(10-k)}{k+1}z_k = \frac{-i(10-k)}{k+1} \binom{10}{k}(-i)^k z_0 = \frac{-i(10-k)}{k+1} \frac{10!}{k!(10-k)!}(-i)^k z_0$ $z_{k+1} = \frac{10!}{(k+1)k!(10-k-1)!} (-i)^{k+1} z_0 = \frac{10!}{(k+1)!(10-(k+1))!} (-i)^{k+1} z_0 = \binom{10}{k+1}(-i)^{k+1} z_0$. So the formula $z_k = \binom{10}{k}(-i)^k z_0$ is correct for $k=0, 1, ..., 10$.

Statement P: The $z_0$ such that $z_0+z_1+z_2+...+z_{10}=2^{10}$ is $(1-i)^{10}$. The sum is $\sum_{k=0}^{10} z_k = \sum_{k=0}^{10} \binom{10}{k}(-i)^k z_0 = z_0 \sum_{k=0}^{10} \binom{10}{k}(-i)^k$. By the binomial theorem, $\sum_{k=0}^{10} \binom{10}{k}(-i)^k = (1+(-i))^{10} = (1-i)^{10}$. So, the sum is $z_0 (1-i)^{10}$. We are given that the sum is $2^{10}$. $z_0 (1-i)^{10} = 2^{10}$. $z_0 = \frac{2^{10}}{(1-i)^{10}} = \left(\frac{2}{1-i}\right)^{10}$. $\frac{2}{1-i} = \frac{2(1+i)}{(1-i)(1+i)} = \frac{2(1+i)}{1-i^2} = \frac{2(1+i)}{1+1} = \frac{2(1+i)}{2} = 1+i$. So, $z_0 = (1+i)^{10}$. Statement P says that $z_0 = (1-i)^{10}$. Since $(1+i)^{10} \ne (1-i)^{10}$ (e.g., $|1+i|^{10} = (\sqrt{2})^{10} = 32$, $|1-i|^{10} = (\sqrt{2})^{10} = 32$, but $\arg((1+i)^{10}) = 10 \frac{\pi}{4} = \frac{5\pi}{2}$ and $\arg((1-i)^{10}) = 10 (-\frac{\pi}{4}) = -\frac{5\pi}{2}$, which are different modulo $2\pi$), statement P is false.

Statement Q: $|z_0|^2+|z_1|^2+|z_2|^2+...+|z_{10}|^2 \ge \frac{(31)^{10}}{10!}$. We need to evaluate this for the specific $z_k$ sequence defined by the sum in statement P. For this sequence, $z_0 = (1+i)^{10}$. $|z_0| = |(1+i)^{10}| = |1+i|^{10} = (\sqrt{1^2+1^2})^{10} = (\sqrt{2})^{10} = 2^5 = 32$. $|z_0|^2 = 32^2 = 1024 = 2^{10}$.

We have $|z_k| = \left|\binom{10}{k}(-i)^k z_0\right| = \binom{10}{k} |-i|^k |z_0| = \binom{10}{k} (1)^k |z_0| = \binom{10}{k} |z_0|$. So, $|z_k|^2 = \left(\binom{10}{k}\right)^2 |z_0|^2$. The sum of squared magnitudes is $\sum_{k=0}^{10} |z_k|^2 = \sum_{k=0}^{10} \left(\binom{10}{k}\right)^2 |z_0|^2 = |z_0|^2 \sum_{k=0}^{10} \left(\binom{10}{k}\right)^2$. Using the identity $\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$, for $n=10$, we have $\sum_{k=0}^{10} \binom{10}{k}^2 = \binom{20}{10}$. The sum of squared magnitudes is $|z_0|^2 \binom{20}{10}$. Substituting $|z_0|^2 = 2^{10}$, the sum is $2^{10} \binom{20}{10}$.

Statement Q is $2^{10} \binom{20}{10} \ge \frac{(31)^{10}}{10!}$. $2^{10} \frac{20!}{10!10!} \ge \frac{(31)^{10}}{10!}$. Multiply by $10!$: $2^{10} \frac{20!}{10!} \ge (31)^{10}$. $\frac{20!}{10!} = 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11$. The inequality is $2^{10} \times (20 \times 19 \times ... \times 11) \ge (31)^{10}$. $(2 \times 20) \times (2 \times 19) \times ... \times (2 \times 11)$ is not correct. The inequality is $2^{10} \times (20 \times 19 \times ... \times 11) \ge 31^{10}$. Let's rewrite the product on the left side: $20 \times 19 \times ... \times 11 = \prod_{j=11}^{20} j$. We are comparing $2^{10} \prod_{j=11}^{20} j$ with $31^{10}$. Let's compare the terms: $20 > 31/2 = 15.5$ $19 > 15.5$ $18 > 15.5$ $17 > 15.5$ $16 > 15.5$ $15 < 15.5$ $14 < 15.5$ ... $11 < 15.5$ Let's rewrite the product $20 \times 19 \times ... \times 11$ as $(15.5+4.5)(15.5+3.5)...(15.5+0.5)(15.5-0.5)...(15.5-4.5)$. The geometric mean of the terms $11, 12, ..., 20$ is $\sqrt[10]{11 \times 12 \times ... \times 20}$. This is greater than the arithmetic mean $\frac{11+...+20}{10} = \frac{155}{10} = 15.5$. So, $\sqrt[10]{11 \times 12 \times ... \times 20} > 15.5$. $\prod_{j=11}^{20} j > (15.5)^{10} = (31/2)^{10} = \frac{31^{10}}{2^{10}}$. Multiplying by $2^{10}$, we get $2^{10} \prod_{j=11}^{20} j > 31^{10}$. So, $2^{10} \frac{20!}{10!} > 31^{10}$. This means $2^{10} \frac{20!}{10!10!} > \frac{31^{10}}{10!}$. The inequality in statement Q is $2^{10} \binom{20}{10} \ge \frac{(31)^{10}}{10!}$. Since we found $2^{10} \binom{20}{10} > \frac{(31)^{10}}{10!}$, the inequality holds. Statement Q is true.

Statement P is false and statement Q is true.