$i{z}^3+z^2-z+i=0\Rightarrow \left|z\right|=$

$i{z}^3+z^2-z+i=0$

$\Rightarrow i{z}^3+i^{2}z+z^2+i=0$

$\Rightarrow iz\left(z^2+i\right)+\left(z^2+i\right)=0$

$\Rightarrow \left(z^2+i\right)+\left(iz+1\right)=0$

$\Rightarrow z^2+i=0$ (or) $iz+1=0$

$\Rightarrow z^2=-i$ (or) $iz=-1\Rightarrow z=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$i$}\right.=i$

$\Rightarrow z^2=1\left(\cos \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+i\sin \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  (or) $z=1\left(\cos \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+i\sin \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

All the roots have modulus as 1. So $\left|z\right|=1$