Question

If $arg\left(z^{1/3}\right)=\frac{1}{2}arg(z^2+\overline{z}{z}^{1/3})$, then find the value of $|z|$.
(Note : $z$ has both real and imaginary components. If $\theta$ is the argument of $z$ , then use $\frac{\theta }{3}$ as the argument of $z^{1/3}$ and $0<\theta <\frac{\pi }{2}$)

Answer 1

We have
$arg\left(z^{1/3}\right)=\frac{1}{2}arg(z^2+\overline{z}{z}^{1/3})$
$\Rightarrow 2arg\left(z^{1/3}\right)=arg(z^2+\overline{z}{z}^{1/3})$

$\Rightarrow arg\left(z^{2/3}\right)=arg(z^2+\overline{z}{z}^{1/3})$ (By Prop.)

$\Rightarrow arg(z^2+\overline{z}{z}^{1/3})-arg\left(z^{2/3}\right)=0$

$\Rightarrow arg\left(\frac{z^2+\overline{z}{z}^{1/3}}{z^{2/3}}\right)=0$ (By Prop.)

$\Rightarrow arg(z^{4/3}+\frac{\overline{z}}{z^{1/3}})=0\Rightarrow z^{4/3}+\frac{\overline{z}}{z^{1/3}}$ is purely real

$\Rightarrow Im(z^{4/3}+\frac{\overline{z}}{z^{1/3}})=0$

$\Rightarrow z^{4/3}+\frac{\overline{z}}{z^{1/3}}=(\overline{z}{)}^{4/3}+\frac{\left(\overline{z}\right)}{(\overline{z}{)}^{1/3}}$

$\Rightarrow z^{4/3}+\frac{\left(\overline{z}\right)(\overline{z}{)}^{1/3}}{|z{|}^{2/3}}=(\overline{z}{)}^{4/3}+\frac{z(z{)}^{1/3}}{|z{|}^{2/3}}$

$\Rightarrow z^{4/3}-(\overline{z}{)}^{4/3}-\frac{1}{|z{|}^{2/3}}((z{)}^{4/3}-(\overline{z}{)}^{4/3})=0$

$\{z^{4/3}-(\overline{z}{)}^{4/3}\}[1-\frac{1}{|z{|}^{2/3}}]=0$

$\therefore |z{|}^{2/3}=1$ $(\because z\ne \overline{z})$

$\therefore |z|=1$
Ans: 1