Question

If \(x_n = \cos \frac{\pi}{3^n} + i \sin \frac{\pi}{3^n}\), then \(x_1 x_2 x_3 ..........\) is equal to

• A. 1
• B. \(-1\)
• C. \(i\)
• D. \(-i\)

Hint:

\(\cos\theta + i\sin\theta = e^{i\theta}\). Sum of geometric series: \(a + ar + ar^2 + .......... = \frac{a}{1-r}\).

Answer 1

The Correct Option is C

To solve the given problem, we start by analyzing the expression for \(x_n\): 

\(x_n = \cos \frac{\pi}{3^n} + i \sin \frac{\pi}{3^n}\)

This expression represents a complex number in polar form. According to Euler's formula, we can rewrite this as:

\(x_n = e^{i \frac{\pi}{3^n}}\)

We are asked to find the product:

\(x_1 x_2 x_3 \ldots\)

Substituting the expression for each \(x_n\), we have:

\(x_1 = e^{i \frac{\pi}{3}}\) 
\(x_2 = e^{i \frac{\pi}{3^2}}\) 
\(x_3 = e^{i \frac{\pi}{3^3}}\) 
\(\ldots\)

Thus, the infinite product becomes:

\(x_1 x_2 x_3 \ldots = e^{i \frac{\pi}{3}} \cdot e^{i \frac{\pi}{3^2}} \cdot e^{i \frac{\pi}{3^3}} \cdot \ldots\)

Using properties of exponents, we can combine these into a single exponential:

\(= e^{i (\frac{\pi}{3} + \frac{\pi}{3^2} + \frac{\pi}{3^3} + \ldots)}\)

This is an infinite geometric series with first term \(a = \frac{\pi}{3}\) and common ratio \(r = \frac{1}{3}\). The sum \(S\) of an infinite geometric series is given by:

\(S = \frac{a}{1 - r}\)

Substituting the values, we find:

\(S = \frac{\frac{\pi}{3}}{1 - \frac{1}{3}} = \frac{\frac{\pi}{3}}{\frac{2}{3}} = \frac{\pi}{2}\)

Therefore, the exponent of \(e\) becomes \(i \frac{\pi}{2}\), giving us the complete expression:

\(e^{i \frac{\pi}{2}}\)

Now, evaluating this, we use the Euler's formula:

\(e^{i \frac{\pi}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\)

Calculating these values, \(\cos \frac{\pi}{2} = 0\) and \(\sin \frac{\pi}{2} = 1\), we find:

\(e^{i \frac{\pi}{2}} = 0 + i \cdot 1 = i\)

Therefore, the product of the series is \(i\).