Let $z = x + iy$. Use modulus and trial with $z = 1$, $z = i$, etc. Try $z = 1$:
\[
iz^3 + z^2 - z + i = i(1)^3 + (1)^2 - 1 + i = i + 1 - 1 + i = 2i \neq 0
\]
Try $z = i$:
\[
iz^3 + z^2 - z + i = i(i^3) + i^2 - i + i = i(-i) + (-1) - i + i = 1 - 1 = 0
\Rightarrow z = i \Rightarrow |z| = 1
\]