= Solution
{c}
The <superficial degree of divergence> of this <scalar electrodynamics three-point vertex> is one. Its <ultraviolet pole> is consequently a local expression at most linear in the external <four-momenta>. The allowed <counterterm Lagrangian>, with the charge-conjugation and Hermiticity properties of <scalar electrodynamics>, contains a single one-derivative <scalar electrodynamics three-point vertex>:
$$
V_{\mathrm{ct}}^\mu(k,k')=-ie\delta_1(k+k')^\mu.
$$
Thus the coefficient $\delta_1$ is independent of the chosen external <four-momenta>. Setting $k'=0$ leaves $-ie\delta_1k^\mu$, so a nonzero generic $k$ still determines that coefficient. It does not determine the full finite <scattering amplitude> at other <four-momenta>.
Keep $m\ne0$ and generic <off shell> $k$ while extracting the <ultraviolet pole>. These choices avoid spurious <infrared divergences>; setting all scales to zero would instead produce <scaleless integrals> that can conceal the <ultraviolet pole>.
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