Question

If \( z_r = \cos \frac{r\alpha}{n^2} + i \sin \frac{r\alpha}{n^2} \), where \( r = 1, 2, 3, ..., n \), then the value of \( \lim_{n \to \infty} z_1 z_2 z_3 ... z_n \) is:

• A. 0
• B. \( e^{\frac{i \alpha}{2}} \)
• C. \( e^{\frac{i \alpha}{2}} \)
• D. \( e^{i \alpha} \)

Hint:

Euler's identity, \( e^{i\theta} = \cos \theta + i \sin \theta \), is useful when working with products of complex numbers in trigonometric form.

Answer 1

The Correct Option is C

Step 1: Expressing \( z_r \) in Exponential Form Given that: \[ z_r = \cos \frac{r\alpha}{n^2} + i \sin \frac{r\alpha}{n^2} \] Using Euler’s formula, we can rewrite it as: \[ z_r = e^{i \frac{r\alpha}{n^2}} \]
Step 2: Computing the Product \( z_1 z_2 z_3 \cdots z_n \) The product is: \[ z_1 z_2 z_3 \cdots z_n = e^{i \frac{\alpha}{n^2}} \cdot e^{i \frac{2\alpha}{n^2}} \cdot e^{i \frac{3\alpha}{n^2}} \cdots e^{i \frac{n\alpha}{n^2}} \] \[ = e^{i \left( \frac{\alpha}{n^2} (1 + 2 + 3 + \dots + n) \right)} \]
Step 3: Evaluating the Summation We use the formula for the sum of the first \( n \) natural numbers: \[ 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \] Thus, the exponent simplifies to: \[ \frac{\alpha}{n^2} \times \frac{n(n+1)}{2} = \frac{\alpha(n+1)}{2n} \]
Step 4: Taking the Limit as \( n \to \infty \) \[ \lim_{n \to \infty} \frac{\alpha(n+1)}{2n} = \frac{\alpha}{2} \] Therefore: \[ \lim_{n \to \infty} z_1 z_2 z_3 \cdots z_n = e^{i \frac{\alpha}{2}} \]
Step 5: Verifying the Correct Option Since we found: \[ \lim_{n \to \infty} z_1 z_2 z_3 \cdots z_n = e^{\frac{i \alpha}{2}} \] This matches option (C) \( e^{\frac{i \alpha}{2}} \).