Question

If the least positive integer \( n \) satisfying the equation \(\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^n = -1\) is \( p \) and the least positive integer \( m \) satisfying the equation \(\left(\frac{1-\sqrt{3}i}{1+\sqrt{3}i}\right)^m = \text{cis}\left(\frac{2\pi}{3}\right)\) is \( q \), then \(\sqrt{p^2 + q^2}\) is equal to:

• A. \(5\)
• B. \(10\)
• C. \( \sqrt{13} \)
• D. \( \sqrt{17} \)

Hint:

When dealing with powers of complex numbers, convert them to polar form \( re^{i\theta} \). Use De Moivre's Theorem \( (re^{i\theta})^n = r^n e^{in\theta} \). Remember that arguments are periodic with \( 2\pi \), so \( \theta = \theta_0 + 2k\pi \) for integer \( k \). For the least positive integer, choose the smallest \( k \) that yields a positive result.

Answer 1

The Correct Option is C

Step 1: Simplify the complex fraction in the first equation.
Let \(z_1 = \sqrt{3}+i\) and \(z_2 = \sqrt{3}-i\). Convert them to polar form \(r(\cos\theta + i\sin\theta) = re^{i\theta}\).
For \(z_1 = \sqrt{3}+i\):
Magnitude \(r_1 = |\sqrt{3}+i| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3+1} = \sqrt{4} = 2\).
Argument \(\theta_1 = \arg(\sqrt{3}+i) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
So, \(z_1 = 2e^{i\frac{\pi}{6}}\).
For \(z_2 = \sqrt{3}-i\):
Magnitude \(r_2 = |\sqrt{3}-i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = \sqrt{4} = 2\).
Argument \(\theta_2 = \arg(\sqrt{3}-i) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}\).
So, \(z_2 = 2e^{-i\frac{\pi}{6}}\).
Now, simplify the ratio: \[ \frac{\sqrt{3}+i}{\sqrt{3}-i} = \frac{2e^{i\frac{\pi}{6}}}{2e^{-i\frac{\pi}{6}}} = e^{i\left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right)} = e^{i\frac{2\pi}{6}} = e^{i\frac{\pi}{3}}. \] The first equation becomes \(\left(e^{i\frac{\pi}{3}}\right)^n = -1\). Using De Moivre's Theorem, \(e^{i\frac{n\pi}{3}} = -1\). We know that \(-1\) in polar form is \(e^{i\pi}\) (or \(e^{i(\pi + 2k\pi)}\) for integer \(k\)). So, we have: \[ e^{i\frac{n\pi}{3}} = e^{i\pi} \] Equating the arguments: \[ \frac{n\pi}{3} = \pi + 2k\pi \] Divide by \(\pi\): \[ \frac{n}{3} = 1 + 2k \] \[ n = 3(1 + 2k) \] For the least positive integer \(n\), we take \(k=0\): \[ n = 3(1 + 0) = 3. \] So, \(p = 3\). 
Step 2: Simplify the complex fraction in the second equation.
Let \(w_1 = 1-\sqrt{3}i\) and \(w_2 = 1+\sqrt{3}i\). Convert them to polar form.
For \(w_1 = 1-\sqrt{3}i\):
Magnitude \(r_1' = |1-\sqrt{3}i| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2\).
Argument \(\theta_1' = \arg(1-\sqrt{3}i) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}\). (Since \(w_1\) is in the 4th quadrant) So, \(w_1 = 2e^{-i\frac{\pi}{3}}\).
For \(w_2 = 1+\sqrt{3}i\):
Magnitude \(r_2' = |1+\sqrt{3}i| = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2\).
Argument \(\theta_2' = \arg(1+\sqrt{3}i) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}\). (Since \(w_2\) is in the 1st quadrant) So, \(w_2 = 2e^{i\frac{\pi}{3}}\). Now, simplify the ratio: \[ \frac{1-\sqrt{3}i}{1+\sqrt{3}i} = \frac{2e^{-i\frac{\pi}{3}}}{2e^{i\frac{\pi}{3}}} = e^{i\left(-\frac{\pi}{3} - \frac{\pi}{3}\right)} = e^{-i\frac{2\pi}{3}}. \] The second equation becomes \(\left(e^{-i\frac{2\pi}{3}}\right)^m = \text{cis}\left(\frac{2\pi}{3}\right).\)
We know that \(\text{cis}\left(\frac{2\pi}{3}\right) = e^{i\frac{2\pi}{3}}\).
Using De Moivre's Theorem, \(e^{-i\frac{2m\pi}{3}} = e^{i\frac{2\pi}{3}}\).
Equating the arguments: \[ -\frac{2m\pi}{3} = \frac{2\pi}{3} + 2k\pi \] Divide by \(\frac{2\pi}{3}\): \[ -m = 1 + 3k \] \[ m = -(1 + 3k) \] For the least positive integer \(m\), we choose \(k\) such that \(1+3k\) is the most negative, which makes \(m\) the smallest positive integer. If \(k=-1\): \[ m = -(1 + 3(-1)) = -(1 - 3) = -(-2) = 2. \] So, \(q = 2\). 
Step 3: Calculate \( \sqrt{p^2 + q^2} \).
We found \(p=3\) and \(q=2\).
Substitute these values into the expression \( \sqrt{p^2 + q^2} \): \[ \sqrt{p^2 + q^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}. \]