Question

The smallest positive integral value of \( n \) such that \[ \left( \frac{1 + \sin \frac{\pi}{8} + i \cos \frac{\pi}{8}}{1 + \sin \frac{\pi}{8} - i \cos \frac{\pi}{8}} \right)^n \] is purely imaginary, is equal to:

• A. 4
• B. 3
• C. 2
• D. 8

Hint:

When solving problems involving complex exponentiation, use the fact that for an expression to be purely imaginary, the real part (cosine term) must be zero. This allows you to solve for the appropriate values of the exponent.

Answer 1

The Correct Option is A

We are given the expression: \[ \frac{1 + \sin \frac{\pi}{8} + i \cos \frac{\pi}{8}}{1 + \sin \frac{\pi}{8} - i \cos \frac{\pi}{8}} \] Let \( z = 1 + \sin \frac{\pi}{8} + i \cos \frac{\pi}{8} \). The conjugate of \( z \) is \( \bar{z} = 1 + \sin \frac{\pi}{8} - i \cos \frac{\pi}{8} \). Thus, we can rewrite the expression as: \[ \frac{z}{\bar{z}} = \frac{1 + \sin \frac{\pi}{8} + i \cos \frac{\pi}{8}}{1 + \sin \frac{\pi}{8} - i \cos \frac{\pi}{8}} \] This expression simplifies as: \[ \frac{z}{\bar{z}} = e^{i 2 \cdot \frac{\pi}{8}} = e^{i \frac{\pi}{4}} \] Step 1: Take the power of the expression Now, we need to find the smallest positive integer \( n \) such that: \[ \left( e^{i \frac{\pi}{4}} \right)^n \] is purely imaginary. Using the properties of complex exponentiation: \[ e^{i n \frac{\pi}{4}} = \cos \left( n \frac{\pi}{4} \right) + i \sin \left( n \frac{\pi}{4} \right) \] For the expression to be purely imaginary, the real part must be zero, i.e., \[ \cos \left( n \frac{\pi}{4} \right) = 0 \] This occurs when \( n \frac{\pi}{4} = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \), or \( n = 2, 6, 10, \dots \). Thus, the smallest positive integer \( n \) such that the expression is purely imaginary is \( n = 4 \). Thus, the correct answer is Option A.