If $z=i(i+\sqrt{2})$,then value of $z^4+4z^3+6z^2+4z$ is

$z=i(i+\sqrt{2})=-1+\sqrt{2}i$

$\Rightarrow (z+1{)}^2=-2\Rightarrow z^2+2z+3=0$

We now divide

$z^4+4z^3+6z^2+4z\text{ by }{z}^2+2z+3$

$z^2+2z+3\sqrt{z^4+4z^3+6z^2+4z}\left[z^2+2z-1\right.$

$\frac{{\mathrm{z}}^4+2z^3+3z^2}{2z^3+3z^2+4z}$

$2z^3+4z^2+6z$

$\therefore z^4+4z^3+6z^2+4z$

$=\left(z^2+2z+3\right)\left(z^2+2z-1\right)+3$

$=0+3=3$