= Solution
{c}
Let the bare fields obey the <wave-function renormalization> definitions $\psi_0=Z_2^{1/2}\psi$, $A_{0\mu}=Z_3^{1/2}A_\mu$, and put $m_0=Z_m m$. Define the <vertex renormalization> factor by $e_0=\mu^{\epsilon/2}eZ_1Z_2^{-1}Z_3^{-1/2}$. Substitution into the bare <QED> <Lagrangian density> gives
$$
\boxed{\mathcal L=Z_2\bar\psi i\not\partial\psi-Z_2Z_m m\bar\psi\psi-\frac{Z_3}{4}F_{\mu\nu}F^{\mu\nu}-e_DZ_1\bar\psi\gamma^\mu\psi A_\mu.}
$$
Equivalently, write $Z_i=1+\delta_i$ and $\delta_m=Z_2Z_m-1$. The <counterterm Lagrangian> is
$$
\mathcal L_{\mathrm{ct}}=\delta_2\bar\psi i\not\partial\psi-\delta_m m\bar\psi\psi-\frac{\delta_3}{4}F_{\mu\nu}F^{\mu\nu}-e_D\delta_1\bar\psi\gamma^\mu\psi A_\mu.
$$
For the subsequent <Feynman gauge> calculation include $-(\partial\cdot A)^2/2$ in the <gauge-fixed action>. The <Ward identity for QED renormalization constants> implies $Z_1=Z_2$ when the <regularization> preserves <gauge invariance>; this is a consequence, rather than an extra independent definition of the <renormalization constants>.
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