You have to solve the linear system
$$
2\begin{pmatrix}
i & -1\\
1 & i
\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=
\begin{pmatrix}0 \\ 0\end{pmatrix}
$$
which becomes $ix_1-x_2=0$. A nonzero solution of this system is thus
$$
\begin{pmatrix}1 \\ i\end{pmatrix}
$$
It's no different from the “real” case.
As Amzoti observes, the fact that your matrix has real coefficients implies that $\lambda_2=\bar\lambda_1$; so if $v=(1~i)^T$ and $\bar{v}=(1~-i)^T$ is the “conjugate” of $v$, you have
$$
\bar{A}\bar{v}=\overline{Av}=\overline{\lambda_1 v}=\bar\lambda_1\bar{v}=\lambda_2\bar{v}
$$
so that $\bar{v}$ is an eigenvector for $\lambda_2$. In other words, any time you find an eigenvector for a complex (non real) eigenvalue of a real matrix, you get for free an eigenvector for the conjugate eigenvalue.
As you have found, the eigenvalues of $A = \begin{bmatrix} 3 & -2 \\ 3 & 1 \end{bmatrix}$ are $\lambda_1 = 2 - \sqrt{5} i$ and $\lambda_2 = 2 + \sqrt{5} i$. And so, we can find the corresponding eigenvectors by solving $(A - \lambda I) = \mathbf{0}$. For example, for $\lambda_1$,
$$\begin{bmatrix} 3 - (2 - \sqrt{5}i) & -2 \\ 3 & 1 - (2 - \sqrt{5}i) \end{bmatrix} \mathbf{x} = \mathbf{0}$$
As you have probably found out, doing row reduction on the above is difficult. However, there's a trick: $(A - \lambda I) = \mathbf{0}$ must have non-trivial solution, which implies that the matrix $(A - \lambda I)$ is not invertible. That means, $(A - \lambda I)$'s rows are linearly dependent. What does that mean? We can pick one of the equivalent linear equations below, because they both describe the same relationship between $x_1$ and $x_2$:
$$3 - (2 - \sqrt{5}i) x_1 - 2 x_2 = 0$$
$$3 x_1 + 1 - (2 - \sqrt{5}i) x_2 = 0$$
Suppose we pick the first one:
$$3 - (2 - \sqrt{5}i) x_1 - 2 x_2 = 0$$
$$1 + \sqrt{5}i x_1 - 2 x_2 = 0$$
$$x_1 = \frac{2 x_2}{1 + \sqrt{5}i}$$
therefore, the eigenvectors corresponding to $\lambda_1$ have the following form:
$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \frac{2 x_2}{1 + \sqrt{5}i} \\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} \frac{2}{1 + \sqrt{5}i} \\ 1 \end{bmatrix}$$
And so, the basis for eigenspace corresponding to $\lambda_1$ is $\Bigg\{\begin{bmatrix} \frac{2}{1 + \sqrt{5}i} \\ 1 \end{bmatrix}\Bigg\}$. You can use the same technique to find the basis for eigenspace corresponding to $\lambda_2$.
Best Answer
The eigenspace relative to $0$ can be deduced from the RREF of the matrix, which is $$ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ This shows there are two free variables; the only equation is $x_1+x_2=0$, so a basis of the eigenspace is obtained by first choosing $x_2=1$ and $x_3=0$, then $x_2=0$ and $x_3=1$: $$ \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$