jmc

Note that $$x^{12}+a{x}^8+b{x}^4+c=p(x^4).$$From the three zeroes, we have $p(x)=(x-(2009+9002\pi i))(x-2009)(x-9002)$. Then, we also have $$p(x^4)=(x^4-(2009+9002\pi i))(x^4-2009)(x^4-9002).$$Let's do each factor case by case:

First, $x^4-(2009+9002\pi i)=0$: Clearly, all the fourth roots are going to be nonreal.

Second, $x^4-2009=0$: The real roots are $\pm \sqrt[4]{2009}$, and there are two nonreal roots.

Third, $x^4-9002=0$: The real roots are $\pm \sqrt[4]{9002}$, and there are two nonreal roots.

Thus, the answer is $4+2+2=\boxed{8}$.