= Solution
The <fundamental theorem of algebra> says that every nonconstant <polynomial> with <complex number> coefficients has a root in the <complex numbers>. Repeated <polynomial division> by a factor of degree one therefore factors a degree-three <polynomial> with <complex number> coefficients into three factors of degree one, with <algebraic multiplicities> included.
For a <matrix> with <complex number> entries $A$ of size three, its <characteristic polynomial> is $\chi_A(t)=\det(tI-A)$, and its characteristic equation is $\chi_A(t)=0$. A <complex number> $\lambda$ is an <eigenvalue> exactly when $\lambda I-A$ is singular, equivalently when its <null space> contains a nonzero <vector>. Thus the three roots of the <characteristic polynomial> are the three <eigenvalues> counted with <algebraic multiplicity>.
For the specified <matrix>, expansion along the first row gives
$$
\chi_A(t)=(t-1)\det\begin{pmatrix}t&-i\\i&t\end{pmatrix}
=(t-1)(t^2-1)=(t-1)^2(t+1).
$$
Hence \b[the <eigenvalues> are $1,1,-1$]. Corresponding <eigenvectors> are
$$
v_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad
v_2=\begin{pmatrix}0\\i\\1\end{pmatrix},\qquad
v_3=\begin{pmatrix}0\\-i\\1\end{pmatrix},
\qquad Av_1=v_1,\quad Av_2=v_2,\quad Av_3=-v_3.
$$
The <matrix> with these three columns has <determinant> $2i\ne0$, so these <eigenvectors> are <linearly independent>. A repeated <eigenvalue> does not in general guarantee two independent <eigenvectors>; here they have been exhibited directly.
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