Differential equation involves complex conjugate root! Find the solution for 
$$\frac{d^4y}{dx^4}+y=0$$
Know that if $y=e^{mx}$
then 
$\frac{d^4y}{dx^4}=m^4e^{mx}$.
The auxiliary equation is $m^4+1=0$. Then
$m^2= \pm  i$.
We have two cases: $m^2= - i$, $m^2= +i$.
Again we further have a total of four cases for complex conjugate roots of the characteristic equation.
$m={-\sqrt {-i}} $, $m={+\sqrt {-i}}$,
$m={-\sqrt i}$, $m={+\sqrt i}$.
We know that $e^{mx}=e^{(a+bi){x}}$, $e^{i \theta}=e^{bx(i)}$
$$e^{(a+bi){x}}=e^{ax} \cos bx+i \sin  bx$$
We have
$b i=\sqrt i$
such that $b i=- i^2 \sqrt i$,
$b=-i \sqrt i$.
Similar reasoning goes for the rest???
I know this may be simple, but just want to confirm the authenticity of this!
 A: Yes, you are on the right track. The solutions of the characteristic equation $z^4=-1$ are the four complex numbers $\frac{\pm 1\pm i}{\sqrt{2}}$. Therefore the general solution of the homogeneous ODE $y^{({iv})}+y=0$ is 
$$y(x)=e^{\frac{x}{\sqrt{2}}}\left(C_1\cos\left(\frac{x}{\sqrt{2}}\right)+C_2\sin\left(\frac{x}{\sqrt{2}}\right)\right)+
e^{-\frac{x}{\sqrt{2}}}\left(C_3\cos\left(\frac{x}{\sqrt{2}}\right)+C_4\sin\left(\frac{x}{\sqrt{2}}\right)\right)$$
where $C_1,C_2,C_3,C_4$ are arbitrary constants.
A: To be fair, I'm not sure whether I understood your question completely or not, but I guess you want to compute the values roots of the auxiliary equations.
As you can see the the roots are the negatives of the fourth roots of unity. To calculate them we can use De Moivre's formula and we have that:
$$-1 = \cos(\pi) + i\cos(\pi)$$
Hence:
$$x_k = \cos\left(\frac{\pi + 2\pi k}{n}\right) + i\sin\left(\frac{\pi +2\pi k}{n}\right)$$
where $k$ varies between $1$ and $n$. Then you can pair the conjugate roots and get a solution, as RobertZ mentioned in his answer.
