Complex-city 2

$i{ z }^{ 3 }+z^{ 2 }-z+i=i{ z }^{ 2 }\left( z-i \right) -\left( z-i \right) =\left( i{ z }^{ 2 }-1 \right) \left( z-i \right) =0$

Case $1:$

$z=i\Rightarrow \left| z \right| =1$

Case $2:$

$i{ z }^{ 2 }=1\Rightarrow { z }^{ 2 }=-i$

$\Rightarrow \left| { z }^{ 2 } \right| =\sqrt { { 0 }^{ 2 }+{ (-1) }^{ 2 } } =1$

$\Rightarrow \left| z \right| =1$ .

In both cases $\left| z \right| =1.\square$

$iz^3+iz^2-z+i=0$

$iz^2(z-i)-(z-i)=0$

Taking $z-i$ common,

$(z-i)(iz^2-1)=0$

Either $z=i$ or $iz^2=1$

$iz^2=1$

$z^2=\frac{1}{i}=-i$

Solving it, we get

$z=\pm\frac{1-i}{\sqrt{2}}$

In both cases, $|z|=1$

Hence $|z|=1$