Question

If $\arg \left(z^{1/3}\right)=\frac{1}{2}\arg (z^2+\overline{z}{z}^{1/3})$, then prove that $\left|z\right|=1$

Answer 1

Given $2\arg \left(z^{1/3}\right)=\arg (z^2+\overline{z}{z}^{1/3})$
or $\arg (z^2+\overline{z}{z}^{1/3})-arg\left(z^{2/3}\right)=0$
$\because n\arg =\arg z^n$
$\arg \left(\frac{z^2+\overline{z}{z}^{1/3}}{z^{2/3}}\right)=0$
$\therefore z^{4/3}+\overline{z}{z}^{-1/3}$ is purely real
$\because \arg z=0\Rightarrow z$ is real
Now we know that if $A$ is real than $A=\overline{A}$
$\therefore z^{4/3}+\frac{\overline{z}}{z^{1/3}}={\left(\overline{z}\right)}^{4/3}+\frac{z}{{\left(\overline{z}\right)}^{1/3}}$
or $z^{4/3}-{\left(\overline{z}\right)}^{4/3}=\frac{z}{{\left(\overline{z}\right)}^{1/3}}-\frac{\overline{z}}{z^{1/3}}=\frac{z^{4/3}-{\left(\overline{z}\right)}^{4/3}}{{\left(z\overline{z}\right)}^{1/3}}$
or $[z^{4/3}-(\overline{z}{)}^{4/3}](1-\frac{1}{{\left|z\right|}^{2/3}})=0$
Since $z\ne \overline{z}$ therefore we have
$1-\frac{1}{{\left|z\right|}^{2/3}}=0$ or ${\left|z\right|}^{2/3}=1$ or $\left|z\right|=1$