QED Lagrangian - real or complex?

I'm confused about the use of complex numbers in the QED Lagrangian: $$\mathcal{L}=\bar{\psi}(i\gamma^{\mu}\partial_{\mu} - m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e\bar{\psi}\gamma^{\mu}A_\mu\psi.$$

Clearly, the Dirac field spinor has complex components. The $$\gamma^\mu$$ matrices involve imaginary numbers.

Is there some algebraic magic that means that $$\mathcal{L}$$ always comes out real, or is it complex? And if it's complex then how does the action $$\int{\mathcal{L}d^4x}$$ come out real? What about $$A^\mu$$ - are its values real, and if so, how does the RHS of the EOM $$\partial_{\nu}F^{\nu\mu}=e\bar{\psi}\gamma^\mu\psi$$ come out real given that the $$\gamma^\mu$$ matrices involve imaginary components?

• As @Toffomat mentioned, try to evaluate what $\mathcal{L}^*$ comes out to be. Clearly some terms are real by construction, such as the terms involving only $A_\mu$s. The other can be found very easily. Apr 16 '21 at 10:57
• OK I take it from these comments that the answer is "algebraic magic makes it real", which I will work my way through. Thanks. Apr 16 '21 at 11:05
• Note that when calculating the complex conjugate, there are serveral places where there exist different conventions (sign of the metric, Hermitean/anti-Hermitean $\gamma$s etc.), so be careful Apr 16 '21 at 11:08

OP's Lagrangian density is up to a total divergence term equal to $${\cal L}~=~\bar{\psi}(\frac{i}{2}\gamma^{\mu}\underbrace{\stackrel{\leftrightarrow}{\partial}_{\!\mu}}_{=\stackrel{\rightarrow}{\partial}_{\mu}-\stackrel{\leftarrow}{\partial}_{\mu}} - m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e\bar{\psi}\gamma^{\mu}A_\mu\psi,\tag{A}$$ which in turn is real. Here we use the conventions $$(\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{B}$$