= Solution
{c}
Only the term proportional to $\alpha$ has an <ultraviolet pole>. Its $k^2$ cancels the massless denominator, leaving the standard <Feynman integral>
$$
J_D=\int\frac{d^Dk}{(2\pi)^D}\frac1{(k^2-m^2+i0)^2}=\frac{i}{(4\pi)^{D/2}}\Gamma(2-D/2)(m^2)^{D/2-2}.
$$
In the present <dimensional regularization> convention, $\Gamma(\epsilon/2)=2/\epsilon+O(1)$ and $\alpha=1+O(\epsilon)$. Therefore
$$
iV^\mu_{\mathrm{loop}}\big|_{\mathrm{UV}}=-ie_D\gamma^\mu\frac{e^2}{8\pi^2\epsilon}.
$$
The <counterterm> contribution is $-ie_D\delta_1\gamma^\mu$. Cancellation of the <ultraviolet pole> fixes
$$
\boxed{Z_1=1+\delta_1=1-\frac{e^2}{8\pi^2\epsilon}+O(e^4)\quad(D=4-\epsilon),}
$$
where the displayed expression uses the <minimal subtraction scheme>; a different <renormalization scheme> may add finite $O(e^2)$ terms. In the frequently used convention $D=4-2\varepsilon$, the same result is $Z_1=1-e^2/(16\pi^2\varepsilon)+O(e^4)$. The sign agrees with $Z_1=Z_2$ from the <Ward identity for QED renormalization constants> in <Feynman gauge>.
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