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
$2^{1/3}\left[\cos \left(\frac{2k\pi +5\pi /3}{2}\right)+i\sin \left(\frac{2k\pi +5\pi /3}{3}\right)\right]$
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
$=\frac{5\pi }{9}+\frac{11\pi }{9}+\frac{17\pi }{9}$
$=\frac{33\pi }{9}=\frac{11\pi }{3}$