Question

Value of \( (1 + i)^8\), where \(i = \sqrt{-1}\), is equal to

• A. 4
• B. 16
• C. \( 4i \)
• D. \( 16i \)

Hint:

Use De Moivre's Theorem for efficient calculations of complex number powers. Convert to polar form, raise to the desired power, and then convert back to rectangular form if needed.

Answer 1

The Correct Option is B

We are tasked with calculating \((1 + i)^8\), where \(i\) is the imaginary unit. We can approach this using polar form of complex numbers. First, express \(1 + i\) in polar form: - The modulus \(r\) of \(1 + i\) is: \[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \] - The argument \(\theta\) is: \[ \theta = \tan^{-1}\left( \frac{1}{1} \right) = \frac{\pi}{4} \] Now, we can rewrite \(1 + i\) as: \[ 1 + i = \sqrt{2} \left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right) \] Using De Moivre's Theorem, which states that for any complex number in polar form \(r(\cos \theta + i \sin \theta)\), the power \(n\) is given by: \[ \left( r(\cos \theta + i \sin \theta) \right)^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right) \] We apply this for \(n = 8\): \[ (1 + i)^8 = (\sqrt{2})^8 \left( \cos\left(8 \times \frac{\pi}{4}\right) + i \sin\left(8 \times \frac{\pi}{4}\right) \right) \] \[ = 16 \left( \cos(2\pi) + i \sin(2\pi) \right) \] Since \(\cos(2\pi) = 1\) and \(\sin(2\pi) = 0\), we get: \[ (1 + i)^8 = 16 \times (1 + 0i) = 16 \] Thus, the correct answer is (B) 16.