Suppose I have a second order differential equation
I can find the complementary solution $y_c$ by finding the roots of the characteristic equation.
Let those roots be $\lambda \pm\mu$
Thus $y_c(t) = C_1e^{t(\lambda + \mu)} + C_2e^{t(\lambda - \mu)}$ where $C_1$ and $C_2$ are real constants
I was taught that this simplifies to $y_c = e^{\lambda}(Acos(\mu t) + Bsin(\mu t))$
I do not understand how the complex part disappears.
$$y_c = C_1e^{t(\lambda + \mu)} + C_2e^{t(\lambda - \mu)} = e^{\lambda t}(C_1e^{\mu t} + C_2e^{-\mu t})$$
$$= e^{\lambda t}(C_1cos(\mu t) +C_1isin(\mu t)+C_2cos(-\mu t) + C_2isin(-\mu t))$$
Since $cos(\mu t) = cos(-\mu t)$ and $sin(\mu t) = -sin(\mu t)$
$$= e^{\lambda t}(C_1cos(\mu t) +C_1isin(\mu t)+C_2cos(-\mu t) + C_2isin(-\mu t))$$
$$= e^{\lambda t}((C_1 + C_2)cos(\mu t) + (C_1-C_2)isin(\mu t))$$
If we let $A = C_1 + C_2$ and $B = C_1 - C_2$ then $$= e^{\lambda t}(Acos(\mu t) + Bisin(\mu t))$$
Notice that there $i$ is still present in the equation. How is the complex part removed?