Solve $z^5 - iz^3+iz^2+1 =0$

Question

Solve for $z$: $$z^5 - iz^3+iz^2+1 =0$$ I've ruled out the approach with euler's representation, de moivre's way led me to $\cos4\phi + \sin3\phi-\sin2\phi +1 = 0$ (with the assumption that the length of $z$ is $1$, which I derived from the fact that $i$ is one of the solutions, which shows trivially through horner's algorithm). Substituting $z$ with $a+ bi$ got me some very ugly numerical solutions for $a$, and as previously mentioned, $i$ is a solution that leads to the equation $$z^4+iz^3+(-1-i)z^2+z+i=0$$I need help.

Answer 1

$$z^5 - iz^3+iz^2+1 =z^3(z^2-i)+i(z^2-i)=(z^3+i)(z^2-i)$$

You should be able to solve it now.

Answer 2

Hint:

$$z^3(z^2-i)+i(z^2-i)=?$$