As you know, real eigenvectors are very important in approximating the phase plane and determine the orientation of the phase curves.
Are the real and imaginary values of complex eigenvectors also effective in plotting the phase plane when the linear system has complex eigenvalues? How?
What is the effect of an imaginary value on the approximate drawing of the phase plane? What is the effect of the real value?
A linear time invariant homogenous dynamical system can be characterised with
$$ \dot{x}(t) = A\,x(t), $$
with initial condition $x(0) \in \mathbb{R}^n$, its evolution $x(t) \in \mathbb{R}^n$ for all $t > 0$ and $A \in \mathbb{R}^{n \times n}$. The eigenvalues and eigenvectors of $A$ allows one to decompose $x(t)$ in modes, that can be described with
$$ x(t) = \sum_{i=1}^n \alpha_i\,v_i\,e^{\lambda_i\,t}, \tag{1} $$
where $(\lambda_i,v_i)$ form an eigenvalue-eigenvector pair of the matrix $A$, such that $A\,v_i = \lambda_i\,v_i$, and $\alpha_i$ is a scalar which also depends on the initial condition $x(0)$.
When the matrix $A$ consists only of real valued elements then complex eigenvalues come in conjugate pairs. In that case the two associated eigenvectors are also complex conjugates from each other. Therefore, those two eigenvalue-eigenvector pairs can also be written as
\begin{align} (\lambda_1, v_1) &= (\rho + \sigma\,i, u + w\,i), \\ (\lambda_2, v_2) &= (\rho - \sigma\,i, u - w\,i), \end{align}
with $i$ the imaginary unit (in case you are more used to notation $j$), the variables $\rho, \sigma \in \mathbb{R}$ and vectors $u, w \in \mathbb{R}^n$. The contribution of these two mode to $(1)$ can be written as
$$ \alpha_1\,v_1\,e^{\lambda_1\,t} + \alpha_2\,v_2\,e^{\lambda_2\,t} = \alpha_1\,(u + w\,i)\,e^{(\rho + \sigma\,i)\,t} + \alpha_2\,(u - w\,i)\,e^{(\rho - \sigma\,i)\,t}. \tag{2} $$
It can be noted that if $A$ and $x(0)$ are only real valued, then $x(t)$ should also always be real valued. In order for $(2)$ to only contribute a real component to $x(t)$ it is required that $\alpha_1$ and $\alpha_2$ are each others complex conjugate. Substituting $\alpha_1 = \beta + \gamma\,i$ and $\alpha_2 = \beta - \gamma\,i$ in $(2)$ and simplify the resulting expression yields
$$ \alpha_1\,v_1\,e^{\lambda_1\,t} + \alpha_2\,v_2\,e^{\lambda_2\,t} = 2\,e^{\rho\,t} \left(\beta \cos(\sigma\,t) - \gamma \sin(\sigma\,t)\right) u - 2\,e^{\rho\,t} \left(\beta \sin(\sigma\,t) + \gamma \cos(\sigma\,t)\right) w. \tag{3} $$
From $(3)$ it can be noted that it always remains in the plane spanned by the vectors $u$ and $w$. Thus, the real and imaginary part of an eigenvector define the plane in which the associated modes oscillate or spiral.
It can be noted that it is not necessary that $v_1$ is the complex conjugate of $v_2$. This is because multiplying an eigenvector by any complex (non-zero) scalar yields another vector that this satisfies $A\,v_i = \lambda_i\,v_i$. It is just that most eigenvalue/eigenvector solvers return the eigenvectors as complex conjugates. However, the plane that is spanned by the real and imaginary vector remains the same. This also means that there isn't any importance to the real vector or imaginary vector, since those can change when multiplying by a complex scalar.