Question

The value of \( (1 + i)^{12} \), where \( i = \sqrt{-1} \), is

• A. \( -64i \)
• B. \( 64i \)
• C. \( 64 \)
• D. \( -64 \)

Hint:

To simplify powers of complex numbers, convert to polar form and use De Moivre's theorem for efficient calculation.

Answer 1

The Correct Option is D

We are given the expression \( (1 + i)^{12} \), where \( i = \sqrt{-1} \). 
 

Step 1: Convert to polar form. 
First, express \( 1 + i \) in polar form. The modulus of \( 1 + i \) is given by: \[ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}. \] The argument \( \theta \) of \( 1 + i \) is: \[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}. \] Thus, we can write: \[ 1 + i = \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right). \]

Step 2: Apply De Moivre's Theorem. 
Using De Moivre's Theorem, we raise the expression to the 12th power: \[ (1 + i)^{12} = \left( \sqrt{2} \right)^{12} \left( \cos\left( 12 \times \frac{\pi}{4} \right) + i \sin\left( 12 \times \frac{\pi}{4} \right) \right). \] This simplifies to: \[ (1 + i)^{12} = 2^6 \left( \cos\left( 3\pi \right) + i \sin\left( 3\pi \right) \right). \] Since \( \cos(3\pi) = -1 \) and \( \sin(3\pi) = 0 \), we have: \[ (1 + i)^{12} = 64(-1 + 0i) = -64. \]