I know that a complex field has twice the number of degrees of freedom of a real field, and that fields (in QFT) aren't observables so we don't really care if they are real.
But why the need for complex fields? Is there stuff that doesn't work unless there's a complex field?
There is no non-trivial one-dimensional representation of $\mathrm{U}(1)$ on a scalar field $\mathbb{R}^4\to\mathbb{R}$, but on complex fields $\mathbb{R}^4\to\mathbb{C}$, we have the one-dimensional "phase" representations by $$\phi\mapsto\mathrm{e}^{e\mathrm{i}\chi}\phi$$ for $e\in\mathbb{Z},\chi\in\mathfrak{u}(1)\cong\mathbb{R}$ for $\mathrm{U}(1)$ parametrized as $\chi\mapsto \mathrm{e}^{\mathrm{i}\chi}$ (the unit circle in the complex plane).
Since $\mathrm{U}(1)$ is the archetypical example of a continuous (gauge) symmetry (think of electromagnetism), complex scalar fields are an important (toy) model in QFT.
Of course, every complex scalar field may equivalently be replaced by two real scalar fields being its real and imaginary part, so they are not actually needed, but using only real fields may complicate the actual calculations and notations immensely.
When switching from a complex scalar $\phi$ to two real ones $\mathrm{Re}(\phi),\mathrm{Im}(\phi)$, we observe that $$ \mathrm{e}^{e\mathrm{i}\chi}\phi = (\cos(e\chi) + \mathrm{i}\sin(e\chi))(\mathrm{Re}(\phi) + \mathrm{i}\ \mathrm{Im}(\phi))$$ and so, writing the real vector $\widetilde{\phi} = \left( \begin{matrix} \phi_1 := \mathrm{Re}(\phi) \\ \phi_2 := \mathrm{Im}(\phi)\end{matrix}\right)$, we see that the complex one-dimensional representation of $\mathrm{U}(1)$ turns into a two-dimensional real one with $$ \widetilde{\phi}\mapsto R_e(\chi)\widetilde{\phi}$$ with the rotation matrix $$ R_e(\chi) := \left(\begin{matrix}\cos(e\chi) & -\sin(e\chi) \\ \sin(e\chi) & \cos(e\chi)\end{matrix}\right)$$ which is now looking more like a representation of the real 2D rotations $\mathrm{SO}(2)$ (the usual one for $e = 1$). As a real representation, this is irreducible (you cannot diagonalize all rotation matrices at once), so you cannot reduce the degrees of freedom and still have a non-trivial representation of $\mathrm{U}(1)\cong\mathrm{SO}(2)$. Two real d.o.f. are the minimum to have some kind of non-trivial continuous symmetry going on, since $\mathrm{U}(1)$ is the simplest Lie group apart from the un-exciting $\mathbb{R},+$.
What type of fields are you using?
If you are working with spinor fields, the representation of Lorentz transformations is complex. So even if the field is real in some reference frame, if you switch to another reference frame it will become complex. There's no way to avoid complex spinor fields.
Actually, you can do without complex fields, at least in some general and important cases, and I don't mean replacing a complex field with two real fields. Schroedinger noted that, in the case of a scalar field interacting with electromagnetic field (the klein-Gordon-Maxwell electrodynamics, or scalar electrodynamics), you can use the so-called unitary gauge, where the scalar field is real. You can also write an equivalent Lagrangian with a real field (please see, e.g., Eq.14 in my article http://akhmeteli.org/akh-prepr-ws-ijqi2.pdf (published in Int'l J. Quantum Information) - the Lagrangian was derived by Takabayashi). What about spinor fields? @Bosoneando, e.g., believes that "There's no way to avoid complex spinor fields". Surprisingly, there is. I showed in http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.) (see also http://arxiv.org/abs/1502.02351) that three out of for complex components of the Dirac spinor in the Dirac equation can be algebraically eliminated in a general case. The remaining component can be made real by a gauge transform.
