I encountered the following question:
Given the operator defined as $\widehat{A} = m\frac{d^2}{dx^2} +mw^2$ with both $m,w \in \mathbb{R}$. What is the kernel?
Then I read the answer and completely got confused, the answer is: $a\cos(wx) +a\sin(wx)$. How can you get to that result? Is it using $e^{ai} = \cos(a) + i\sin(a)$? Thanks for the clarification.
First off the question is a bit unclear: the operator is on which space? A reasonable assumption would be (twice, or more) differentiable real valued functions $\mathbb{R} \to \mathbb{R}$. Let's call this space $C$. Also we should assume $m$ is nonzero (think of $m$ as mass, for example).
A function $f \in C$ is in $\ker(\hat{A})$ if (and only if) $\hat{A}(f) = 0$. Using our definition of $\hat{A}$: $$ m\frac{d^2}{dx^2}f + m\omega^2 f = 0, $$ so we have the second order differential equation $f'' + \omega^2 f = 0$. We can solve using the characteristic polynomial: $$ r^2 + \omega^2 = 0, $$ which has complex roots $r = \pm i\omega$. The general form of the solution $f(x) = Ae^{rx}$ is thus $f(x) = Ae^{i\omega x}$. Using Euler's formula (as you mentioned): $$ f(x) = a \cos(\omega x) + b \sin(\omega x), $$ as desired ($a$ and $b$ are undetermined constants).