There is given matrix $A$, where z is a complex number. Find how many different complex number there is, that $\det(A^{4}) = 16$.
$A =\begin{bmatrix} z & z+1 & 1\\ 1 & z & z+1 \\ z+1 & 1 & z \end{bmatrix}$
I have calculated that $det(A^4)$ is equal to $(z(z+2)(z-\frac{1}{2}))^4$. And there are four roots of $16$, $2, -2, 2i, -2i$. So let the $p(z)=z(z+2)(z-\frac{1}{2})$. How can I tell how many different roots have polynomials $p(z) - 2 = 0,p(z) + 2 = 0,p(z) - 2i = 0,p(z) + 2i = 0$?
We have $\det(A^4)=16$ if and only if $$ 0=16z^{12} + 64z^9 + 96z^6 + 64z^3=16z^3(z^6 + 2z^3 + 2)(z^3 + 2). $$ So we only have to count the number of complex roots from here. So we have $12$ complex roots. Note that $\det(A^4)=\det(A)^4$.