Q13/082) if iz^3 + z^2 – z + i = 0 ( i is square root of -1) show that |z| = 1

iz^3 + z^2 – z + i

= iz^2( z-i) – (z-i)

= (iz^2 – 1)(z –i)

So z – i = 0 => |z| = 1

or iz^2 = 1 => z^2 = i => |z|^2 = |i| = 1 => |z| = 1