If ${\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3\in \mathrm{C}$ and $\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=\left|{\mathrm{z}}_3\right|=1,{\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3=1$ and ${\mathrm{z}}_1{\mathrm{z}}_2{\mathrm{z}}_3=1$ such that $\mathrm{Im}\left({\mathrm{z}}_1\right)<\mathrm{Im}\left({\mathrm{z}}_2\right)<\mathrm{Im}\left({\mathrm{z}}_3\right)$, then $\left|{\mathrm{z}}_1+{\mathrm{z}}_2^2+{\mathrm{z}}_3^3\right|=$

$\sum _{ }{\mathrm{z}}_1{\mathrm{z}}_2={\mathrm{z}}_1{\mathrm{z}}_2{\mathrm{z}}_3\left(\frac{1}{{\mathrm{z}}_1}+\frac{1}{{\mathrm{z}}_2}+\frac{1}{{\mathrm{z}}_3}\right)$
$={\overline{z}}_1+{\overline{z}}_2+{\overline{z}}_3=1$
so,${\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3$ are roots of the equation ${\mathrm{t}}^3-1\cdot {\mathrm{t}}^2+1\cdot \mathrm{t}-1=0$
i.e.,$(\mathrm{t}-1)\left({\mathrm{t}}^2+1\right)=0\Rightarrow {\mathrm{z}}_1=-\mathrm{i},{\mathrm{z}}_2=1,{\mathrm{z}}_3=\mathrm{i}$
$\therefore \left|-\mathrm{i}+1+{\mathrm{i}}^3\right|=|1-2\mathrm{i}|=\sqrt{5}$