Question

One of the \(5^{\text{th}}\) roots of \(\omega\) is

• A. \(\operatorname{cis}\left(\dfrac{4\pi}{3}\right)\)
• B. \(\operatorname{cis}\left(\dfrac{7\pi}{15}\right)\)
• C. \(\operatorname{cis}\left(\dfrac{11\pi}{15}\right)\)
• D. \(\operatorname{cis}\left(\dfrac{9\pi}{15}\right)\)

Hint:

The \(n^{\text{th}}\) roots of \[ \operatorname{cis}\theta \] are \[ \boxed{ \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right), \qquad k=0,1,\ldots,n-1. } \] Always use this formula to find the roots of a complex number in polar form.

Answer 1

The Correct Option is A

Step 1: Express \(\omega\) in polar form. A cube root of unity is \[ \omega=\operatorname{cis}\left(\frac{2\pi}{3}\right). \]

Step 2: Find the fifth roots of \(\omega\). If \[ z^5=\omega, \] then the fifth roots are \[ z=\operatorname{cis}\left(\frac{\frac{2\pi}{3}+2k\pi}{5}\right), \qquad k=0,1,2,3,4. \] Therefore, \[ z=\operatorname{cis}\left(\frac{2\pi}{15}+\frac{2k\pi}{5}\right). \] Substituting \(k=0,1,2,3,4\), \[ \operatorname{cis}\left(\frac{2\pi}{15}\right),\; \operatorname{cis}\left(\frac{8\pi}{15}\right),\; \operatorname{cis}\left(\frac{14\pi}{15}\right),\; \operatorname{cis}\left(\frac{20\pi}{15}\right),\; \operatorname{cis}\left(\frac{26\pi}{15}\right). \] Since \[ \frac{20\pi}{15}=\frac{4\pi}{3}, \] one of the fifth roots is \[ \boxed{\operatorname{cis}\left(\frac{4\pi}{3}\right).} \] Hence, \[ \boxed{(A)} \] is the correct answer.