Given $p,q$ of the form $4k+3$, we can define complex numbers mod $p$ or mod $q$. The form guarantees that $-1$ has no square root there, so we are working with a quadratic extension of $GF(p)$ and $GF(q)$, i.e.
$$GF(p^2) \simeq GF(p)[i]/(i^2+1)$$
and
$$GF(q^2) \simeq GF(q)[i]/(i^2+1).$$
Combining them, we can define RSA over $(\mathbb{Z}/n\mathbb{Z})[i]/(i^2+1)$, $n=pq$, i.e. complex numbers where real and imaginary parts are defined mod $n$. The arithmetic (multiplication, addition, etc.) is done similarly to complex numbers. E.g.
$$(a+bi)(c+di)\equiv (ac-bd) + (ad+bc)i \pmod{n}$$
(values stored as pairs $(a,b)$ and $(c,d)$).
The order of the multiplicative group is $\mathrm{lcm}(p^2-1, q^2-1)$, so we need
$$e\cdot d \equiv 1 \pmod{\mathrm{lcm}(p^2-1, q^2-1)}.$$
Of course, factoring $n$ breaks the scheme, so there's no increase in security.
Another possible issue is that complex numbers have the (squared) norm, which is multiplicative and easy to compute without knowing $p$ and $q$:
$$N(a+bi) \equiv a^2+b^2 \pmod{n},$$
$$N(c) \equiv N(m^e) \equiv N(m)^e \pmod{n}.$$
Since the norm lives in $GF(p)\times GF(q)$, we reduce to basic RSA mod $n$. If an attacker can somehow break RSA mod $n$, the attacker recovers then the norm of the message.
Summary:
It is possible to define RSA over complex numbers modulo primes, however the advantages of doing this are not clear, as the security does not increase compared to basic RSA, and efficiency drops a bit.
I don't see any other significant attacks on this scheme, I would be glad to know if there are any.
Upd:
Instead of $\sqrt{-1}$ we can use any other $\sqrt{d}$ that does not lie in the base field, i.e. work with $\mathbb{Z}[\sqrt{d}]/n\mathbb{Z}$. This was done in a variant of OSS signatures. (OSS was broken but only because it's weak itself, RSA defined over quadratic integers should be ok.)