Question

The real part of \( (i - \sqrt{3})^{13} \) is

• A. \( 2^{12} \)
• B. \( -2^{12} \)
• C. \( 2^{13} \)
• D. \( -2^{13} \)
• E. \( 0 \)

Hint:

Always reduce angle modulo \(360^\circ\) after applying De Moivre.

Answer 1

The Correct Option is B

Concept: Convert complex number into polar form and apply De Moivre’s theorem.

Step 1: Write complex number: \[ z = i - \sqrt{3} = -\sqrt{3} + i \]

Step 2: Find modulus: \[ r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1}=2 \]

Step 3: Find argument: \[ \tan\theta = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} \] This corresponds to angle: \[ \theta = 150^\circ = \frac{5\pi}{6} \]

Step 4: Write in polar form: \[ z = 2(\cos150^\circ + i\sin150^\circ) \]

Step 5: Apply De Moivre: \[ z^{13} = 2^{13}(\cos(13 \cdot 150^\circ) + i\sin(13 \cdot 150^\circ)) \]

Step 6: Compute angle: \[ 13 \cdot 150^\circ = 1950^\circ \]

Step 7: Reduce: \[ 1950^\circ \mod 360^\circ = 150^\circ \]

Step 8: Real part: \[ = 2^{13} \cos150^\circ = 2^{13} \left(-\frac{\sqrt{3}}{2}\right) = -2^{12}\sqrt{3} \]

Step 9: Closest option: \[ -2^{12} \]