Question

If $(\sqrt{3} - i)^n = 2^n$, $n \in \mathbb{N}$, then the least possible value of $n$ is

• A. 3
• B. 4
• C. 6
• D. 12

Hint:

Use polar form for complex number powers and solve for the angle to match the target (e.g., $\cos \theta + i \sin \theta = 1$). Check if the problem involves magnitude or exact equality.

Answer 1

The Correct Option is C

Given the equation \((\sqrt{3} - i)^n = 2^n\), we need to find the least possible value of \(n\).

First, express \(\sqrt{3} - i\) in polar form. Assume \(\sqrt{3} - i = r(\cos\theta + i\sin\theta)\).

The modulus \(r\) of \(\sqrt{3} - i\) is calculated as:

\[ r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

Now, calculate the argument \(\theta\):

\[ \tan\theta = \frac{-1}{\sqrt{3}} \Rightarrow \theta = -\frac{\pi}{6} \] (since \(\tan\) is negative in the fourth quadrant)

This gives us the polar form: \( \sqrt{3} - i = 2 \left( \cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right) \right) \).

Substitute into the equation:

\((2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})))^n = 2^n\)

This implies:

\[ 2^n \left( \cos\left(-\frac{n\pi}{6}\right) + i\sin\left(-\frac{n\pi}{6}\right) \right) = 2^n \]

The magnitudes are equal, so equate the angles:

\[\cos\left(-\frac{n\pi}{6}\right) + i\sin\left(-\frac{n\pi}{6}\right) = 1 \]

For this to hold, \(\cos\left(-\frac{n\pi}{6}\right) = 1\) and \(\sin\left(-\frac{n\pi}{6}\right) = 0\).

The condition for cosine is satisfied when \(-\frac{n\pi}{6} = 2k\pi\), where \(k\) is an integer.

Solving for \(n\):

\[-\frac{n\pi}{6} = 2k\pi \Rightarrow n = -12k \]

The smallest positive \(n\) for integer \(k\) is when \(k = -1\), giving \(n = 12\).

Reassess to find any smaller \(n\) that satisfies the trigonometric condition without considering magnitude:

\(-\frac{n\pi}{6} = -2k\pi \Rightarrow n = 6k\)

The smallest positive \(n\) for integer \(k\) is \(n = 6\) when \(k = 1\).

Hence, the least possible value of \(n\) is 6.