Question

Least positive argument of the 4th root of the complex number $2-i\sqrt{12}$ is

• A. $\pi/6$
• B. $5\pi/12$
• C. $7\pi/12$
• D. $11\pi/12$

Answer 1

Answer: (B)

$5\pi/12$

Answer 2

Write the number in polar form:

$$2 - i\sqrt{12} = 4\left(\cos\left(-\frac{\pi}{3}\right)+ i\sin\left(-\frac{\pi}{3}\right)\right)$$

For the fourth roots, the argument is:

$$\theta_k = \frac{-\pi/3 + 2\pi k}{4}, \quad k=0,1,2,3.$$

For $k=0$: $\theta_0 = -\pi/12$ (negative).

For $k=1$: $\theta_1 = \frac{-\pi/3 + 2\pi}{4} = \frac{5\pi}{12}$ (least positive).