Complexity! (5)

$\begin{aligned} iz^3 + z^2 - z + i & = 0 \\ z^2(iz + 1) - z + i & = 0 \\ \frac {z^2}i (-z + i) - z + i & = 0 \\ (i - z)\left( \frac {z^2}i + 1\right) & = 0 \\ \frac {(i-z)(z^2+i)}i & = 0 \\ (i-z)(z^2+i) & = 0 \end{aligned}$

$\implies z = \begin{cases} i & \implies |z| = 1 \\ \pm \sqrt{-i} & \implies |z| = 1 \end{cases}$

Therefore, $2016|z| = \boxed{2016}$

$R(z) = iz^3+z^2-z+i=0\\ R(i) = i(i^3) + i^2 - i + i = 1 - 1 + 0 = 0 \\ \text{Therefore, z=i is a solution to R(z).} \\ \text{On either - dividing R(z) by (z-i) or using synthetic division, you get the quotient as} \\ Q(z) = iz^2 + 0z^2 - 1 iz^2 - 1\\ \text{So other zeros would be, factorizing, Q(z)} \\ Q(z) = 0 \\ iz^2 - 1 = 0 \\ z = \pm\sqrt[]{\dfrac{1}{i}} \\ \text{Therefore for all 3 values of 'z', we get the absolute as} \\ \large z = i = 0 + 1i \\ |z| = \sqrt[]{0^2 + 1^2} = 1 \\ \large z = + \sqrt[]{\dfrac{1}{i}} = i\sqrt[]{i} = 0 + i\sqrt[]{i}\\ |z| = \sqrt[]{0^2 + (i\sqrt[]{i})^2} = 1 \\ \large z = -i\sqrt[]{i} \\ |z| = 1$