Obtaining eigenvalues of a complex matrix Let
$$A = \begin{bmatrix} 0 & i+1 & 0 \\ i & 0 &2 -i \\ 2-i & 0 & i \end{bmatrix}$$
and eigenvalues, eigenvectors are asked.

*

*by the frequently used method  ,we get the characteristic equation as $$\lambda^{3} - i (1 + \lambda )\lambda + \lambda - 8 = 0$$ no way to solve this equation by hand. By using some WEB sites 3 different complex roots are obtained, all with  "ugly" coefficient like  $(-1.0463-1.6363i),\dots$ Then by using these eigenvalues, almost impossible to get the eigenvectors by hand.


*Diagonalizing matrix $A$ by row operations and picking  up the diagonal entries.
I've done this too. I've obtained "nice" eigenvalues but they do not satisfy the original characteristic equation given above. So there is some inconsistance.
What I've missed here? Any help  will be appreciated.
(This was an exam question, so it is supposed to be solved in a limited time by hand)

The characteristic polynomial is given by
$$
\chi(t)=t^3 - it^2 + ( 1 - i)t - 8
$$
and the roots $z_1,z_2,z_3$ satisfy
$$
z_1+z_2+z_3=i,\; z_1z_2z_3=8.
$$
So we have $z_3=i-z_1-z_2$ and $z_1z_2(i-z_1-z_2)=8$.
This gives a quadratic equation.
