Question

One of the 15th roots of \( -1 \) is:

• A. \( \text{cis} 0 \)
• B. \( \text{cis} \frac{14\pi}{15} \)
• C. \( \text{cis} \frac{13\pi}{15} \)
• D. \( \text{cis} \frac{8\pi}{15} \)

Hint:

To find the \( n \)-th roots of a complex number, use the formula \( z_k = \text{cis} \left( \frac{\theta + 2k\pi}{n} \right) \) and substitute appropriate values of \( k \).

Answer 1

The Correct Option is C

The general formula for the \( n \)-th roots of a complex number \( z = r \text{cis} \theta \) is given by: \[ z_k = \text{cis} \left( \frac{\theta + 2k\pi}{n} \right), \quad k = 0, 1, 2, \dots, n-1 \] For \( -1 \), we have \( r = 1 \) and \( \theta = \pi \). We need to find one of the 15th roots of \( -1 \), so we apply the formula with \( n = 15 \): \[ z_k = \text{cis} \left( \frac{\pi + 2k\pi}{15} \right) \] For \( k = 6 \), we get: \[ z_6 = \text{cis} \left( \frac{\pi + 2(6)\pi}{15} \right) = \text{cis} \left( \frac{13\pi}{15} \right) \] Thus, the correct answer is \( \text{cis} \frac{13\pi}{15} \), which corresponds to option (3).

Answer 2

Given:
We are to find one of the 15th roots of \( -1 \).

Step 1: Express \( -1 \) in polar (cis) form
\[ -1 = \text{cis} \pi = \cos \pi + i\sin \pi \]

Step 2: Use De Moivre’s Theorem
The \( n \)th roots of a complex number \( z = r \text{cis} \theta \) are given by:
\[ z_k = r^{1/n} \text{cis} \left( \frac{\theta + 2k\pi}{n} \right), \quad k = 0, 1, 2, \dots, n-1 \]
Here, \( r = 1 \), \( \theta = \pi \), \( n = 15 \)

So the 15th roots of \( -1 \) are:
\[ z_k = \text{cis} \left( \frac{\pi + 2k\pi}{15} \right) = \text{cis} \left( \frac{(2k + 1)\pi}{15} \right) \]
For \( k = 6 \), we get:
\[ z_6 = \text{cis} \left( \frac{(2 \cdot 6 + 1)\pi}{15} \right) = \text{cis} \left( \frac{13\pi}{15} \right) \]

Final Answer:
One of the 15th roots of \( -1 \) is \( \text{cis} \frac{13\pi}{15} \).