A physical scenario in which a spring is hanging with a weight of mass $m$ on it yelds the following differential equation for the position $z(t)$: $$mz'' = mg - f(z - z_0)$$ In order to solve for $z$ we first need to solve the homogenous equation: $$mz'' + fz = 0$$
Assume $z(t) = e^{\lambda t}$: $$\lambda^2 m e^{\lambda t} + fe^{\lambda t} = 0$$ $$\lambda_{1,2} = \mp i\sqrt{\frac{f}{m}}$$ Having two eigenvalues, the general solution of the homogenous differential equation is a linear combination of the two: $$z_{\rm hom.}(t) = c_1e^{-i\sqrt{\frac{f}{m}}t} + c_2e^{i\sqrt{\frac{f}{m}}t}$$ We're in the context of physics and this is an oscillation, so we want to transform this result using trigonometric functions to make it more explicit. According to the solution supplied to me this is how it goes: \begin{align} z_{\rm hom.}(t) &= c_1e^{i\sqrt{\frac{f}{m}}t} + c_2e^{-i\sqrt{\frac{f}{m}}t} \\ &= c_1\left(\cos\left(\sqrt{\frac{f}{m}}t\right) + i\sin\left(\sqrt{\frac{f}{m}}t\right)\right) + c_2\left(\cos\left(\sqrt{\frac{f}{m}}t\right) - i\sin\left(\sqrt{\frac{f}{m}}t\right)\right) \\ &= (c_1+c_2)\cos\left(\sqrt{\frac{f}{m}}t\right) + (c_1-c_2)i\sin\left(\sqrt{\frac{f}{m}}t\right) \end{align} And here it gets funny: let $c_1+c_2 = A\cos(\phi)$ and $(c_1-c_2)i = A\sin(\phi)$: \begin{align} z_{\rm hom.}(t) &= A\cos(\phi)\cos\left(\sqrt{\frac{f}{m}}t\right) + A\sin(\phi)\sin\left(\sqrt{\frac{f}{m}}t\right) \\ &= A\cos\left(\sqrt{\frac{f}{m}}t - \phi\right) \end{align} So, my question is: why can we say that $c_1+c_2 = A\cos(\phi)$ and $(c_1-c_2)i = A\sin(\phi)$, with the same $A$ and $\phi$? It is awfully convenient, but how come $c_1$ and $c_2$ are related just this way? I'm guessing it has to do with the fact that the two eigenvalues $\lambda_{1,2}$ are complex conjugates, but exactly how escapes me. Any help would be appreciated!
\cos
rather thancos
when typing your formulas. $\endgroup$ – zipirovich Mar 22 '17 at 15:29