The descending order of the values of

${\mathrm{z}}_1={\left[\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}\right]}^6, {\mathrm{z}}_2={\left[\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}\right]}^4$

${\mathrm{z}}_3={\left[\frac{\sqrt{3}}{4}+\frac{\mathrm{i}}{4}\right]}^6$

$\begin{array}{rcl}{\mathrm{z}}_1& =& {\left(\frac{1}{2}+\frac{\mathrm{i}\sqrt{3}}{2}\right)}^6={\left(\cos \frac{\mathrm{\pi }}{3}+\mathrm{i} \sin \frac{\mathrm{\pi }}{3}\right)}^6\\ & =& \cos 2\mathrm{\pi }+\mathrm{i} \sin 2\mathrm{\pi }=1+\mathrm{i}\left(0\right)=1\end{array}$

$\begin{array}{rcl}{\mathrm{z}}_2& =& {\left(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}\right)}^4\\ & =& {\left(\cos \frac{\mathrm{\pi }}{4}+\mathrm{i} \sin \frac{\mathrm{\pi }}{4}\right)}^4\\ & =& \cos \mathrm{\pi }+\mathrm{i} \sin \mathrm{\pi }=-1+\mathrm{i}\left(0\right)=-1\end{array}$

$\begin{array}{rcl}{\mathrm{z}}_3& =& {\left(\frac{\sqrt{3}}{4}+\frac{\mathrm{i}}{4}\right)}^6\\ & =& \frac{1}{2^6}{\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)}^6\\ & =& \frac{1}{2^6}{\left(\cos \frac{\mathrm{\pi }}{6}+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}\right)}^6\\ & =& \frac{1}{2^6}\left(\cos \mathrm{\pi }+\mathrm{i} \sin \mathrm{\pi }\right)=\frac{1}{2^6}(-1)=\frac{-1}{2^6}\\ {\mathrm{z}}_1& >& {\mathrm{z}}_2>{\mathrm{z}}_3\end{array}$