Question

The sum of the least positive arguments of the distinct cube roots of the complex number $(1-i \sqrt{3})$ is

• A. $\frac{5 \pi}{3}$
• B. $\frac{17 \pi}{3}$
• C. $\frac{23 \pi}{3}$
• D. $\frac{11 \pi}{3}$

Answer 1

Answer: (D)

Let $z=1-i \sqrt{3}$
positive argument of $z$ is $2 \pi-\frac{\pi}{3}=\frac{5 \pi}{3}$
$\therefore z=\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right)$
Cube root of $z$ is
$2^{1 / 3}\left[\cos \left(\frac{2 k \pi+5 \pi / 3}{2}\right)+i \sin \left(\frac{2 k \pi+5 \pi / 3}{3}\right)\right]$
where $k=0,1,2$
$\therefore k =0 \arg \left(z^{1 / 3}\right)=\frac{5 \pi}{9}$
$k =1 \arg \left(z^{1 / 3}\right)=\frac{11 \pi}{9}$
$k =2 \arg \left(z^{1 / 3}\right)=\frac{17 \pi}{9}$
Sum of positive argument
$=\frac{5 \pi}{9}+\frac{11 \pi}{9}+\frac{17 \pi}{9}$
$=\frac{33 \pi}{9}=\frac{11 \pi}{3}$