Question

The value of \(z = \frac{(-1 + i)^4}{(\sqrt{3} + i)^{10}}\) is:

• A. \(2^{-9}(-1 - i\sqrt{3})\)
• B. \(2^{8}(1 - i\sqrt{3})\)
• C. \(2^{-8}(-1 + i\sqrt{3})\)
• D. \(2^{9}(-1 + i\sqrt{3})\)

Hint:

Exam Tip:
When dealing with complex numbers with large powers:

• Always convert to polar form.
• Use De Moivre's theorem.
• Simplify using trigonometric identities.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves complex numbers and their powers. We need to simplify the expression using the polar form (De Moivre's theorem).

Step 2: Key Formula or Approach:
For a complex number \(z = r(\cos \theta + i \sin \theta)\), \(z^n = r^n(\cos n\theta + i \sin n\theta)\).
We need to express both \((-1 + i)\) and \((\sqrt{3} + i)\) in polar form.

Step 3: Detailed Explanation:
First, find the polar form of \((-1 + i)\): \[ r_1 = \sqrt{(-1)^2 + 1^2} = \sqrt{2}, \quad \theta_1 = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \quad \text{(since the point is in the second quadrant)} \] So, \((-1 + i) = \sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)\).
Now, find the polar form of \((\sqrt{3} + i)\): \[ r_2 = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2, \quad \theta_2 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \] So, \((\sqrt{3} + i) = 2\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\).
Now, compute \(z\): \[ z = \frac{(-1 + i)^4}{(\sqrt{3} + i)^{10}} = \frac{(\sqrt{2})^4 \left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)^4}{2^{10} \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)^{10}} \] Using De Moivre's theorem: \[ z = \frac{(\sqrt{2})^4 \left(\cos \frac{3\pi}{4} \cdot 4 + i \sin \frac{3\pi}{4} \cdot 4\right)}{2^{10} \left(\cos \frac{\pi}{6} \cdot 10 + i \sin \frac{\pi}{6} \cdot 10\right)} \] \[ z = \frac{4 \left(\cos 3\pi + i \sin 3\pi\right)}{1024 \left(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6}\right)} \] \[ z = \frac{4}{1024} \cdot \frac{\cos 3\pi + i \sin 3\pi}{\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3}} \] \(\cos 3\pi = -1\), \(\sin 3\pi = 0\), so numerator is \(-1\).
\(\cos \frac{5\pi}{3} = \frac{1}{2}\), \(\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}\).
So, \[ z = \frac{4}{1024} \cdot \frac{-1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} = \frac{1}{256} \cdot \frac{-1}{\frac{1 - i\sqrt{3}}{2}} = \frac{1}{256} \cdot \frac{-2}{1 - i\sqrt{3}} \] \[ z = \frac{-2}{256(1 - i\sqrt{3})} = \frac{-1}{128(1 - i\sqrt{3})} = \frac{-1}{128} \cdot \frac{1 + i\sqrt{3}}{1 + 3} = \frac{-1}{128} \cdot \frac{1 + i\sqrt{3}}{4} = \frac{-(1 + i\sqrt{3})}{512} \] \[ z = \frac{-1 - i\sqrt{3}}{512} = 2^{-9}(-1 - i\sqrt{3}) \] Since \(512 = 2^9\).

Step 4: Final Answer:
The value is \(2^{-9}(-1 - i\sqrt{3})\). Therefore, option (A) is correct.