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: Express \( z_r \) in Exponential Form Given:\[z_r = \cos \frac{r\alpha}{n^2} + i \sin \frac{r\alpha}{n^2}\]By Euler's formula, this is equivalent to:\[z_r = e^{i \frac{r\alpha}{n^2}}\]
Step 2: Calculate the Product \( z_1 z_2 z_3 \cdots z_n \) The product is:\[z_1 z_2 z_3 \cdots z_n = e^{i \frac{\alpha}{n^2}} \cdot e^{i \frac{2\alpha}{n^2}} \cdot e^{i \frac{3\alpha}{n^2}} \cdots e^{i \frac{n\alpha}{n^2}}\]This simplifies to:\[= e^{i \left( \frac{\alpha}{n^2} (1 + 2 + 3 + \dots + n) \right)}\]
Step 3: Evaluate the Summation Using the sum of the first \( n \) natural numbers formula:\[1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}\]The exponent becomes:\[\frac{\alpha}{n^2} \times \frac{n(n+1)}{2} = \frac{\alpha(n+1)}{2n}\]
Step 4: Compute the Limit as \( n \to \infty \) The limit of the exponent is:\[\lim_{n \to \infty} \frac{\alpha(n+1)}{2n} = \frac{\alpha}{2}\]Therefore, the limit of the product is:\[\lim_{n \to \infty} z_1 z_2 z_3 \cdots z_n = e^{i \frac{\alpha}{2}}\]
Step 5: Confirm the Correct Option Our result:\[\lim_{n \to \infty} z_1 z_2 z_3 \cdots z_n = e^{\frac{i \alpha}{2}}\]Corresponds to option (C) \( e^{\frac{i \alpha}{2}} \).