Fixed points are most elegantly treated in a Hamiltonian formalism so this is what I am going to use here (I will comment briefly how does this translate to Lagrangian formalism in the end).
Consider a Hamiltonian $H(\theta_1,\theta_2,p_1,p_2)$ and a fixed point (equilibrium) at $\theta_{10},\theta_{20},p_{10},p_{20}$, which is defined by $\dot{\theta}_1 = \dot{\theta}_2 = \dot{p}_1 = \dot{p}_2 = 0$. Considering Hamilton's equations this can be also stated as $dH=0$ at the equilibrium point in phase space. If we then take an initial condition with a small perturbation from the equilibrium $\theta_{10} + \delta \theta_1,\theta_{20}+ \delta \theta_2,p_{10}+ \delta p_1,p_{20}+ \delta p_2$ we obtain a set of equations for this perturbation of the form
$$ \frac{d}{dt}
\begin{pmatrix}
\delta \theta_1 \\ \delta p_1 \\ \delta \theta_2 \\ \delta p_2
\end{pmatrix}
=
\begin{pmatrix}
\frac{\partial^2 H}{\partial p_1 \partial \theta_1} & \frac{\partial^2 H}{\partial p_1^2 } & \frac{\partial^2 H}{\partial p_1 \partial \theta_2} & \frac{\partial^2 H}{\partial p_1 \partial p_2}
\\
-\frac{\partial^2 H}{\partial \theta_1^2} & -\frac{\partial^2 H}{\partial \theta_1 \partial p_1 } & -\frac{\partial^2 H}{\partial \theta_1 \partial \theta_2} & -\frac{\partial^2 H}{\partial \theta_1 \partial p_2}
\\
\frac{\partial^2 H}{\partial p_2 \partial \theta_1} & \frac{\partial^2 H}{\partial p_2 \partial p_1 } & \frac{\partial^2 H}{\partial p_2 \partial \theta_2} & \frac{\partial^2 H}{\partial p_2^2}
\\
-\frac{\partial^2 H}{\partial \theta_2 \partial \theta_1} & -\frac{\partial^2 H}{\partial \theta_2 \partial p_1 } & -\frac{\partial^2 H}{\partial \theta_2^2} & -\frac{\partial^2 H}{\partial \theta_2 \partial p_2}
\end{pmatrix} \begin{pmatrix}
\delta \theta_1 \\ \delta p_1 \\ \delta \theta_2 \\ \delta p_2
\end{pmatrix}$$
That looks a little bit overwhelming, but you can substitute your favorite Hamiltonian in there and you will see typically only a few of the components of this perturbation matrix are non-zero. The differential equation above is a linear differential equation of the form $\dot{\mathbf{v}} = \mathbf{A} \mathbf{v}$ whose solution you can find in almost every linear algebra book.
The full set of solutions is found by an Ansatz of the form $\mathbf{v}(t) = \mathbf{v}_0 e^{\lambda t}$. From that you get that the solutions have to have $\mathbf{v}_0$ an eigenvector of the matrix $\mathbf{A}$ with a possibly complex eigenvalue $\lambda$. There will be generally four eigenvalues and four corresponding linearly independent eigenvectors of $\mathbf{A}$.
Now, the mathematical theory of classical mechanics tells us that due to the "symplectic structure" of classical mechanics, the eigenvalues will be grouped in something called a loxodromic set. A loxodromic set of numbers in the complex plane is a set of numbers which is symmetric under reflections with respect to both the real and imaginary axes. For instance, this is a loxodromic set of four numbers
$$\{\alpha + i\beta, -\alpha + i\beta, \alpha - i \beta, - \alpha - i \beta \}$$
where $\alpha,\beta$ are real numbers. A fixed point with this kind of set of eigenvalues corresponds to a system with a runaway oscillation $e^{(\alpha+i\beta) t}$. This case can happen only in two degrees of freedom and higher, you will not get intuition for it from one degree of freedom! One particular case where this can happen are rotating systems, where the particle is forced to corotate and oscillate by some attractive potential but also spirals out due to the centrifugal force.
One of the other possibilities for the eigenvalues is a "degenerate" loxodromic set of the form
$$\{\alpha,-\alpha, i\beta,-i\beta\}$$
with $\alpha,\beta$ again real. These correspond to a system where a perturbation in one direction causes a stable oscillation whereas in the other direction it is unstable and the trajectory starts drifting away. The loxodromic symmetry of the eigenvalues corresponds to the reversibility of the classical-mechanical system. This means that for every motion in one general direction, you can also find the inverse in some sense; for every $e^{\alpha t}$ runaway, there is an $e^{-\alpha t}$ slow approach to the equilibrium (this corresponds to a particle which has just the right energy to reach a potential maximum and is slowly crawling up towards it).
Now for your question: your Ansatz $\theta_1 = A e^{i\omega t}, \theta_2 = B e^{i\omega t}$ corresponds to the more general Ansatz $\mathbf{v}(t) = \mathbf{v}_0 e^{\lambda t}$ in the Hamiltonian formalism and should also give you a loxodromic set of frequencies.
But you should now see that if we have an initial condition at a general position and velocity close to the equilibrium, it will be a linear combination of the basis of all the four linearly independent eigendirections - independently of whether the loxodromic set is degenerate or non-degenerate. (Remember that you have to give four numbers $\theta_1,\theta_2,\dot{\theta}_1, \dot{\theta}_2$ as your initial condition!!)
If your initial condition (and dynamical system) is real, the loxodromy of the set ensures that your initial conditions will actually combine in a real basis of evolutions written for a non-degenerate set as follows
$$\frac{1}{2i} (e^{(\alpha + i \beta)t} - e^{(\alpha - i \beta})t), \frac{1}{2} (e^{(\alpha + i \beta)t} + e^{(\alpha - i \beta})t), \frac{1}{2i} (e^{(-\alpha + i \beta)t} - e^{(-\alpha - i \beta})t), \frac{1}{2} (e^{(-\alpha + i \beta)t} + e^{(-\alpha - i \beta)t})$$
and similarly for the degenerate sets. So even though the frequencies can get mixed-complex (note purely imaginary or real) for a system with two or more degrees of freedom, the symmetries of the mechanical system save the day and you still get proper non-complexified evolution.
