The ascending order of the value of

${\mathrm{z}}_1={\left(1+\mathrm{i}\right)}^4, {\mathrm{z}}_2={\left(\sqrt{3}+\mathrm{i}\right)}^{12}, {\mathrm{z}}_3={\left(1+\mathrm{i}\sqrt{3}\right)}^9$

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

$\begin{array}{rcl}{\mathrm{z}}_2& =& {\left(\sqrt{3}+\mathrm{i}\right)}^{12}\\ & =& 2^{12}{\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)}^{12}\\ & =& 2^{12}{\left(\cos \frac{\mathrm{\pi }}{6}+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}\right)}^{12}\\ & =& 2^{12}\left(\cos 2\mathrm{\pi }+\mathrm{i} \sin 2\mathrm{\pi }\right)=2^{12}\left(1\right)=2^{12}\end{array}$

$\begin{array}{rcl}{\mathrm{z}}_3& =& {\left(1+\mathrm{i}\sqrt{3}\right)}^9\\ & =& 2^9{\left(\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2}\right)}^9\\ & =& 2^9{\left(\cos \frac{\mathrm{\pi }}{3}+\mathrm{i} \sin \frac{\mathrm{\pi }}{3}\right)}^9\\ & =& 2^9\left(\cos 3\mathrm{\pi }+\mathrm{i} \sin 3\mathrm{\pi }\right)\\ & =& 2^9(-1+\mathrm{i}(0\left)\right)\\ & =& -2^9\end{array}$

$\therefore$  Ascending order is ${\mathrm{z}}_3<{\mathrm{z}}_1<{\mathrm{z}}_2$