I'm trying to understand the math notes detailing the way to obtain the complex solution:
If the problem is homogeneous we have :
$$ay''(t) + by'(t) + cy(t) = 0$$
to know all solutions for this use the ansatz
$$y(t) = e^{\lambda t}$$
and plug it into the ODE to find
$$ay''(t) + by'(t) + cy(t) = a \lambda^2e^{\lambda t} + b\lambda e^{\lambda t} + ce^{\lambda t} = 0$$
This problem can be solved using the characteristic equation:
$$a\lambda^2 + b\lambda +c = 0$$
Dependeing on the value of the discriminant $D = b^2 - 4ac$ we find different types of solutions:
$1.$ $D < 0$: $\lambda_1 \neq \lambda_2$ and both real.
Then we have two linearly independent solutions
$$y_1(t) = e^{\lambda_1t} \quad \text{and}\quad y_2(t) = e^{\lambda_2t}$$ and thus the general solution is
$$y(t) = c_1y_1(t) + c_2y_2(t) = c_1e^{\lambda_1t} + c_2e^{\lambda_2t}$$
$2.$ $D = 0$: $\lambda_1 = \lambda_2 \in \mathbb{R}$
One solution is immediate from the previous case
$$y_1(t) = e^{\lambda_1t}$$
One can verify that the second solution is given by
$$y_2(t) = te^{\lambda_1t}$$ Thus the general solution is
$$y(t) = c_1y_1(t) + c_2y_2(t) = c_1e^{\lambda_1t} (c_1 +tc_2)$$
$3.$ $D < 0$: $\lambda_{1,2} = \alpha \pm i\beta$ with $\alpha$, $\beta \in \mathbb{R}$ and $\beta \neq 0$
Using the idea from the first case we end up with complex sotluions
$$u_1(t) =e^{\lambda_1t} = e^{(\alpha + i \beta)t} = e^{\alpha t}(\cos(\beta t) + i \sin(\beta t)) \tag{1}\label{eq1}$$ $$ u_2(t) =e^{\lambda_2t} = e^{(\alpha + i \beta)t} = e^{\alpha t}(\cos(\beta t) - i \sin(\beta t)) \tag{2}\label{eq2}$$
My Question:
I guess the statement $(1)$ and $(2)$ have to do with the properties of complex numbers. However can you give me a link to understand those $2$ statements.