Question

If $z=-1$, then principal value of the $\arg \left(z^{2/3}\right)$ is/are

• A. $0$
• B. $\frac{2\pi }{3}$
• C. $-\frac{2\pi }{3}$
• D. $\pi$

Answer 1

Answer: (A), (B), (C)

$-\frac{2\pi }{3}$

Given $z=-1\Rightarrow \theta =\pi$
$\Rightarrow z^2=1\Rightarrow (z^2{)}^{1/3}=1,\omega ,{\omega }^2$
$\therefore \arg \left(z^{2/3}\right)=\arg \left(1\right),\arg \left(\omega \right),\arg \left({\omega }^2\right)\Rightarrow \arg \left(z^{2/3}\right)=0,\frac{2\pi }{3},-\frac{2\pi }{3}$