Question

\( (1+\sqrt{3}i)^6 - (\sqrt{3}+i)^6 = \)

• A. \( 0 \)
• B. \( 32 \)
• C. \( 64 \)
• D. \( 128 \)

Hint:

When dealing with powers of complex numbers, converting them to polar form \( r(\cos\theta + i\sin\theta) \) and applying De Moivre's Theorem \( (r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta)) \) is generally the most efficient method. It avoids tedious binomial expansions for high powers. Remember the common angles and their sine/cosine values for quick conversions.

Answer 1

The Correct Option is D

We need to evaluate the expression \( (1+\sqrt{3}i)^6 - (\sqrt{3}+i)^6 \). We will use De Moivre's Theorem, which states that for a complex number \(z = r(\cos\theta + i\sin\theta)\), \(z^n = r^n(\cos(n\theta) + i\sin(n\theta))\). First, let's convert \(1+\sqrt{3}i\) into polar form. Let \(z_1 = 1+\sqrt{3}i\). The modulus \(r_1 = |1+\sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2\). The argument \(\theta_1\) is such that \(\cos\theta_1 = \frac{1}{2}\) and \(\sin\theta_1 = \frac{\sqrt{3}}{2}\). So, \( \theta_1 = \frac{\pi}{3} \). Thus, \( 1+\sqrt{3}i = 2\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right) \). Now, calculate \( (1+\sqrt{3}i)^6 \): Using De Moivre's Theorem: \( (1+\sqrt{3}i)^6 = 2^6\left(\cos\left(6 \cdot \frac{\pi}{3}\right) + i\sin\left(6 \cdot \frac{\pi}{3}\right)\right) \) \( = 64\left(\cos(2\pi) + i\sin(2\pi)\right) \) \( = 64(1 + i \cdot 0) \) \( = 64 \) Next, let's convert \( \sqrt{3}+i \) into polar form. Let \(z_2 = \sqrt{3}+i\). The modulus \(r_2 = |\sqrt{3}+i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2\). The argument \(\theta_2\) is such that \(\cos\theta_2 = \frac{\sqrt{3}}{2}\) and \(\sin\theta_2 = \frac{1}{2}\). So, \( \theta_2 = \frac{\pi}{6} \). Thus, \( \sqrt{3}+i = 2\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) \). Now, calculate \( (\sqrt{3}+i)^6 \): Using De Moivre's Theorem: \( (\sqrt{3}+i)^6 = 2^6\left(\cos\left(6 \cdot \frac{\pi}{6}\right) + i\sin\left(6 \cdot \frac{\pi}{6}\right)\right) \) \( = 64\left(\cos(\pi) + i\sin(\pi)\right) \) \( = 64(-1 + i \cdot 0) \) \( = -64 \) Finally, subtract the two results: \( (1+\sqrt{3}i)^6 - (\sqrt{3}+i)^6 = 64 - (-64) \) \( = 64 + 64 \) \( = 128 \)