Question

The value of \[ \left( \frac{1 + \cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)} {1 + \cos\left(\frac{\pi}{12}\right) - i\sin\left(\frac{\pi}{12}\right)} \right)^{72} \] is equal to: 

• A. \( 0 \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( \frac{1}{2} \)
• E. \( -\frac{1}{2} \)

Hint:

Any expression of the form \( \left( \frac{1+\text{cis}\theta}{1+\text{cis}(-\theta)} \right) \) is simply \( \text{cis}\theta \). Recognizing this identity saves you from tedious rationalization of complex denominators.

Answer 1

The Correct Option is C

Concept: For a complex number \( z = \cos \theta + i \sin \theta \), the expression \( \frac{1+z}{1+\bar{z}} \) simplifies to \( z \). This is because \( \bar{z} = \frac{1}{z} \) for any complex number with unit modulus. Applying De Moivre's Theorem, \( ( \cos \theta + i \sin \theta )^n = \cos(n\theta) + i \sin(n\theta) \).

Step 1: Simplifying the internal fraction.
Let \( z = \cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right) \). Then the expression inside the parentheses is: \[ \frac{1+z}{1+\bar{z}} = \frac{1+z}{1+\frac{1}{z}} = \frac{1+z}{\frac{z+1}{z}} = z \] So, the entire expression becomes \( z^{72} \).

Step 2: Applying De Moivre's Theorem.
\[ z^{72} = \left( \cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right) \right)^{72} \] \[ = \cos\left( 72 \times \frac{\pi}{12} \right) + i\sin\left( 72 \times \frac{\pi}{12} \right) \] \[ = \cos(6\pi) + i\sin(6\pi) \] Since \( \cos(6\pi) = 1 \) and \( \sin(6\pi) = 0 \), the value is \( 1 \).