General solution of the ordinary differential equation $(D^4+D^2+1)y=0$ $(D^4+D^2+1)y=0$ where $D=\frac{d}{dx}$
$$D^4+D^2+1 =0 \Rightarrow D^4 + D^2 + \frac14 +\frac{{3}}{4} = 0 \Rightarrow (D^2 + \frac12) = \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D^2 = \frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D= \pm \sqrt{\frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2}} = \pm \sqrt{e^{\pm(2/3)\pi\iota}} = \pm e^{\pm(1/3)\pi\iota}  $$ 
Is there a way to simplify this?
 A: Your equation is
$$x^4+x^2+1=0$$
Let $y=x^2$, then we have that:
$$y^2+y+1=0$$
And the solution is $y=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$, or in exponential form: $y_1=e^{\frac{2}{3}i\pi}$ and $y_2=e^{-\frac{2}{3}i\pi}$. And from $x^2=y$, we get that $x_{12}=\exp\left(\frac{\frac{2}{3}i\pi+2ni \pi}{2}\right)$ and $x_{34}=\exp\left(\frac{-\frac{2}{3}i\pi+2ni \pi}{2}\right)$ for $n=0$ and $n=1$. So the roots are:
$$x_1=\exp\left(\frac{1}{3}i \pi\right)$$
$$x_2=\exp\left(\frac{4}{3}i \pi\right)$$
$$x_3=\exp\left(-\frac{1}{3}i \pi\right)=\exp\left(\frac{5}{3}i \pi\right)$$
$$x_4=\exp\left(\frac{2}{3}i \pi\right)$$
A: The roots of $\lambda^4+\lambda^2+1$ are $$ \lambda_1=\frac{1}{2} + i\frac{\sqrt{3}}{2}\\ \lambda_2=\frac{1}{2} - i\frac{\sqrt{3}}{2}\\ \lambda_3=-\frac{1}{2} + i\frac{\sqrt{3}}{2}\\ \lambda_4=-\frac{1}{2} - i\frac{\sqrt{3}}{2} $$ 
Thus, the solution to the ODE is 
$$ y(x) = \sum_{k=1}^4 a_k\exp(x\lambda_k) $$
If we write $\mu_{1,2} = \pm\Re(\lambda_1), \sigma = \Im(\lambda_1)$, then
$$ y(x) = \sum_{k=1}^2 [b_k\cos(x\sigma) + c_k\sin(x\sigma)]\exp(x\mu_k) $$
