Roots of Unity Properties Matching List

To find the correct matching between List I and List II, let us analyze each statement individually.

We are given $z_k = \cos\left(\frac{2k\pi}{10}\right) + i \sin\left(\frac{2k\pi}{10}\right) = e^{i\frac{2k\pi}{10}}$ for $k = 1, 2, \dots, 9$. These are the nine non-real $10^{\text{th}}$ roots of unity (excluding $z_0 = 1$).

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Analysis of Statement P:

For any $z_k = e^{i\frac{2k\pi}{10}}$ with $k \in \{1, 2, \dots, 9\}$, we choose $j = 10 - k$.

Since $1 \le k \le 9$, it follows that $1 \le j \le 9$, so $z_j$ belongs to the given set of complex numbers. Calculating the product $z_k \cdot z_j$: $z_k \cdot z_j = e^{i\frac{2k\pi}{10}} \cdot e^{i\frac{2(10-k)\pi}{10}} = e^{i\frac{20\pi}{10}} = e^{i 2\pi} = 1$

Thus, for each $z_k$, there exists a $z_j$ such that $z_k \cdot z_j = 1$.

• P matches with 1 (True).

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Analysis of Statement Q:

For any $k \in \{1, 2, \dots, 9\}$, the equation $z_1 \cdot z = z_k$ gives: $z = \frac{z_k}{z_1} = \frac{e^{i\frac{2k\pi}{10}}}{e^{i\frac{2\pi}{10}}} = e^{i\frac{2(k-1)\pi}{10}}$

Since $z_1 \neq 0$, a unique complex solution $z = e^{i\frac{2(k-1)\pi}{10}} \in \mathbb{C}$ always exists for every $k$. Hence, the claim that there exists a $k$ for which no solution exists is false.

• Q matches with 2 (False).

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Analysis of Statement R:

The $10^{\text{th}}$ roots of unity satisfy the polynomial equation $z^{10} - 1 = 0$. Factoring this polynomial using its roots $1, z_1, z_2, \dots, z_9$: $z^{10} - 1 = (z - 1)(z - z_1)(z - z_2)\dots(z - z_9)$

Dividing both sides by $(z - 1)$: $z^9 + z^8 + \dots + z + 1 = (z - z_1)(z - z_2)\dots(z - z_9)$

Substituting $z = 1$: $1 + 1 + \dots + 1 = 10 = (1 - z_1)(1 - z_2)\dots(1 - z_9)$

Taking the absolute value on both sides: $|1 - z_1||1 - z_2|\dots|1 - z_9| = 10$

Therefore: $\frac{|1 - z_1||1 - z_2|\dots|1 - z_9|}{10} = \frac{10}{10} = 1$

• R matches with 3 (1).

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Analysis of Statement S:

The sum of all $10^{\text{th}}$ roots of unity is zero: $1 + \sum_{k=1}^9 z_k = 0 \implies \sum_{k=1}^9 z_k = -1$

Taking the real part on both sides: $\text{Re}\left(\sum_{k=1}^9 z_k\right) = \sum_{k=1}^9 \cos\left(\frac{2k\pi}{10}\right) = -1$

Now, evaluating the expression in statement S: $1 - \sum_{k=1}^9 \cos\left(\frac{2k\pi}{10}\right) = 1 - (-1) = 2$

• S matches with 4 (2).

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Conclusion:

Matching List I with List II yields: $\mathbf{P \to 1, \quad Q \to 2, \quad R \to 3, \quad S \to 4}$

This corresponds to Option C.