If $\arg \left({\mathrm{z}}^{3/8}\right)=(1/2)\arg \left({\mathrm{z}}^2+\overline{\mathrm{z}}{\mathrm{z}}^{1/2}\right)$ then which of the following is (are) true ?

$\begin{array}{l}2{\mathrm{argz}}^{3/8}=\arg \left({\mathrm{z}}^2+\overline{\mathrm{z}}\sqrt{\mathrm{z}}\right)\\ {\mathrm{argz}}^{3/4}=\arg \left({\mathrm{z}}^2+\overline{\mathrm{z}}\sqrt{\mathrm{z}}\right)\Rightarrow \arg \left({\mathrm{z}}^2+\overline{\mathrm{z}}\sqrt{\mathrm{z}}\right)-{\mathrm{argz}}^{3/4}=0\end{array}$

$\Rightarrow \frac{{\mathrm{z}}^2+\overline{\mathrm{z}}\sqrt{\mathrm{z}}}{{\mathrm{z}}^{3/4}}\text{ is real ⇒Imaginary part of}\left({\text{ z}}^{5/4}\text{+}\frac{\overline{\text{z}}}{z^{1/4}}\right)\text{=0}$

$$\begin{gathered} \Rightarrow {\mathrm{z}}^{\frac{5}{4}}+\frac{\overline{\mathrm{z}}}{{\mathrm{z}}^{\frac{1}{4}}}=\left({\overline{\mathrm{z}}}^{\frac{5}{4}}+\frac{\mathrm{z}}{\overline{\mathrm{z}}\frac{1}{4}}\right) \\ \Rightarrow {\mathrm{z}}^{\frac{5}{4}}-{\overline{\mathrm{z}}}^{\frac{5}{4}}=\frac{\mathrm{z}}{\overline{\mathrm{z}}\frac{1}{4}}-\frac{\overline{\mathrm{z}}}{{\mathrm{z}}^{\frac{1}{4}}} \\ \Rightarrow {\mathrm{z}}^{\frac{5}{4}}-{\overline{\mathrm{z}}}^{\frac{5}{4}}=\frac{{\mathrm{z}}^{\frac{5}{4}}-{\overline{\mathrm{z}}}^{\frac{5}{4}}}{{\left|\mathrm{z}\right|}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \Rightarrow \begin{array}{l}\Rightarrow \left({\mathrm{z}}^{\frac{5}{4}}-\overline{\mathrm{z}}\frac{\overline{5}}{4}\right)\left(1-\frac{1}{\left|z\right|\frac{1}{2}}\right)=0\\ \text{ thus, }z=\overline{z},\left|z\right|=1\\ \end{array} \end{gathered}$$