Question :
For $z=a+bi$, $z\in \mathbb{C}$, find the complete set of values of $Arg(z)$ if the principle argument of $Arg(z^3)$ is in the second quadrant.
My Working :
If $Arg (z^3)$ is in the second quadrant $\theta=\frac\pi2\to\pi$
Let $|z| = x$
$z^3=x^3cis\theta$
$z=xcis\frac\theta3$
$Arg(z)=\frac\pi6\to\frac\pi3$
Was just wondering if someone could confirm that answer and working is correct and if not where I went wrong?
If we are given a complex number $w=r\,e^{i\phi}\ne0$ there are three complex numbers $z$ satisfying $z^3=w$, namely $$z_-={\root 3\of r}\,e^{i(\phi-2\pi)/3}, \quad z_0={\root 3\of r}\,e^{i\phi/3},\quad z_+={\root 3\of r}\,e^{i(\phi+2\pi)/3}\ .$$ Since we are told that ${\pi\over2}<\phi<\pi$ these three numbers have (principal values of their) arguments in the following ranges: $$-{\pi\over2}<{\rm Arg}(z_-)<-{\pi\over3},\qquad {\pi\over6}<{\rm Arg}(z_0)<{\pi\over3},\qquad {5\pi\over6}<{\rm Arg}(z_+)<\pi\ .$$