Confused by ODE $f''(x)=\lambda f(x).$ So the solutions to the characteristic equation is $r_{1,2}=\pm\sqrt{\lambda}.$ We thus have three cases to consider in order to find the solution of $f''=\lambda f.$ However for this question, the only relevant one is when the roots become complex, that is
Case 1: $\lambda < 0.$

Theorem: a second order homogenous ODE of the form $ay''+by'+cy=0$
  with complex roots $r_{1,2}=a\pm bi$ for the characteristic equation
  have solutions of the form
$$y(x)=e^{ax}(A\cos(bx)+B\sin(bx)).\tag1$$

My prof writes that in my case, I have solutions of the form
$$f(x)=A\cos\left(\sqrt{|\lambda|}x\right)+B\sin\left(\sqrt{|\lambda|}x\right).\tag{2}$$
Questions:
1) How is this possible if we dont not know what the complex number is? That is, she doesn't have $a$ and $b$ in order to plug it into $(1).$
2) Why has my prof omitted the $e^{ax}$ factor?
3) Why do we need the absolute value bars around the $\lambda$:s in $(2)?$

1) First, there's a conlict of notation : the $a$ in $ay''+by'+c=0$ is not the same $a$ appearing in $r_{\pm}=a\pm bi$.
That being said, $f''=\lambda f\Longleftrightarrow f''-\lambda f=0$. Its characteristic equation is $r^2-\lambda=0$, which can be easily solved.
2) She didn't. It's just that in this case, the solutions to the characteristic equation have their real parts equal to $0$.
3) Cos' the solution to the characteristic equations are $r_{\pm}=\pm i|\lambda|$.
