Basic tools
For $2 \times 2$ matrices the characteristic polynomial is:
$$
p(\lambda) = \lambda^{2} - \lambda\, \text{tr }\mathbf{A} + \det \mathbf{A}
$$
The roots of this function are the eigenvalues, $\lambda_{k}$, k=1,2$.
The eigenvectors solve the eigenvalue equation
$$
\mathbf{A} u_{k} = \lambda_{k} u_{k}
$$
Case 1
$$
\mathbf{A} =
%
\left(
\begin{array}{cc}
i & 1 \\
0 & -1+i \\
\end{array}
\right)
$$
The trace and determinant are
$$
\text{tr } \mathbf{A} = -1 + 2 i, \qquad \det \mathbf{A} = -1 - i
$$
The characteristic polynomial is
$$
p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i)
$$
The roots of this polynomial are the eigenvalues:
$$ \lambda \left( \mathbf{A} \right) = \left\{
i, -1 + i
\right\}
$$
First eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cc}
i & 1 \\
0 & -1+i \\
\end{array}
\right) -
\left( -1 + i \right)
%
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\left(
\begin{array}{cc}
u_{x} + u_{y}\\
0 \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right)
%
\end{align}
$$
The result is
$$
u_{1} =
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
=
\alpha
\left(
\begin{array}{r}
1 \\
-1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
Second eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cr}
0 & 1 \\
0 & -1 \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\end{align}
$$
The result is
$$
u_{2} =
\alpha
\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
Case 2
$$
\mathbf{A} =
%
\left(
\begin{array}{cr}
\cos (\theta ) & -\sin (\theta ) \\
\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right)
$$
The trace and determinant are
$$
\text{tr } \mathbf{A} = 2 \cos \theta, \qquad \det \mathbf{A} = \cos^{2} \theta + \sin^{\theta} = 1
$$
The characteristic polynomial is
$$
p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i)
$$
The roots of this polynomial are the eigenvalues:
$$ \lambda \left( \mathbf{A} \right) = \left\{
\cos \theta -i \sin \theta ,\cos \theta +i \sin \theta
\right\}
$$
First eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cr}
i \sin \theta & -\sin \theta \\
\sin \theta & i \sin \theta \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\left(
\begin{array}{cc}
-u_{y} \sin \theta +i u_{x} \sin \theta \\
u_{x} \sin \theta +i u_{y} \sin \theta
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right)
%
\end{align}
$$
The result is
$$
u_{1} =
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
=
\alpha
\left(
\begin{array}{r}
i \\
-1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
Second eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{rr}
-i \sin \theta & -\sin \theta \\
\sin \theta & -i \sin \theta \\
\end{array}
\right)
%
\left(
\begin{array}{c}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\end{align}
$$
The result is
$$
u_{2} =
\alpha
\left(
\begin{array}{c}
i \\
1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$