# Missing a factor in the conserved current equation for QFT

I am currently reading some notes in QFT and I came across the conserved current equation for a complex scalar field that has a transformation given by $$\hat{\phi}\longrightarrow\hat{\phi}+i\theta\hat{\phi}$$ The notes said that the current is: $$J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a=i\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi-i\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi\dagger)}\delta\phi\dagger$$ but I thought that $\delta\hat{\phi}=i\theta\hat{\phi}$, then should there be a factor of $\theta$ on all terms?

In the complex scalar field Lagrangian, you have two different dynamic fields: $\psi$ and $\psi^{\dagger}$. The conserved Noether current in this case is defined as $J^{\mu} = \sum_{i = 1}^2 \Pi^{\mu}_i \Delta \psi_i$, where the indices 1 and 2 denote the two different fields.
The symmetry you want is $\psi \rightarrow \psi e^{i\theta}$, and therefore $\psi^{\dagger} \rightarrow \psi^{\dagger} e^{-i\theta}$. If you define $\delta\psi_i \equiv \Delta \psi \delta \theta$, you get that $\Delta \psi = i \psi$ and $\Delta \psi^{\dagger} = -i \psi^{\dagger}$, and you remove the $\delta \theta$ from the definition of the current. Replacing this into the current gives you $$J^{\mu} = i(\Pi^{\mu}_{\psi^{\dagger}}\psi - \Pi^{\mu}_{\psi}\psi^{\dagger})$$ which is what you obtained.
The factor $\delta \theta$ (with $\theta$ a continuous parameter) is not included in the definition of the conserved current because it is just a variation of a parameter and it is irrelevant in the conservation equation $\partial_{\mu}J^{\mu} = 0$. If you were to include it (using $\delta \psi$ instead of $\Delta \psi$ in the definition of $J^{\mu}$), you would get $(\partial_{\mu}J^{\mu})\delta\theta = 0$, which just implies $\partial_{\mu}J^{\mu} = 0$ for non-zero variations. So you just eliminate $\delta \theta$ out of the definition of $J^{\mu}$ because it's irrelevant to the physics.
The definitions on your question are somewhat ambiguous. Remember that $\delta \psi = \frac{\partial \psi}{\partial \theta} \delta \theta$, so you could never have factors of $\theta$ in the definition ($\frac{\partial}{\partial \theta} (e^{i \theta}) = ie^{i \theta}$, NOT $i \theta e^{i \theta}$), only factors of $\delta \theta$