If $i{z}^3+z^2–z+i=0$  then the value of   $\left|z\right|=------$

${\mathrm{iz}}^3+{\mathrm{z}}^2-\mathrm{z}+\mathrm{i}=0$
By substituting z = i in the equation, we get 0 = 0. 
Hence z – i is a factor of ${\mathrm{iz}}^3+{\mathrm{z}}^2-\mathrm{z}+\mathrm{i}$
$\Rightarrow {\mathrm{iz}}^2(\mathrm{z}-\mathrm{i})-1(\mathrm{z}-\mathrm{i})=0\Rightarrow \left({\mathrm{iz}}^2-1\right)(\mathrm{z}-\mathrm{i})=0$
Either ${\mathrm{iz}}^2-1=0\text{ or }\mathrm{z}-\mathrm{i}=0$
When $\mathrm{z}-\mathrm{i}=0,\mathrm{z}=\mathrm{i}$
$\therefore \left|\mathrm{z}\right|=|0+\mathrm{i}\cdot 1|=\sqrt{0^2+1^2}=1\text{. }$
When ${\mathrm{iz}}^2-1=0,{\mathrm{z}}^2=\frac{1}{\mathrm{i}}=-\mathrm{i}$
$\therefore \left|z^2\right|=|0-\underset{\_}{\underset{\_}{1.1}}|=\sqrt{0^2+(-1{)}^2}=1\therefore \left|z^2\right|=1\text{ or }\left|z\right|=1$
$\therefore$ In any case we have   $\left|z\right|=1$