Charge Renormalization in Peskin (Chapter 7.5)

In Chapter 7.5 in Peskin and Schroeder, the authors define the physical charge in eq. (7.76),
$$text{(physical charge)}=sqrt{Z_3}cdottext{(bare charge)}$$
where $dfrac{1}{1-Pi(0)}equiv Z_3$. Here,
$$Pi^{munu}(q)=(q^2g^{munu}-q^mu q^nu)Pi(q^2)$$
$iPi^{munu}(q)$ being the sum of all 1PI insertions into the photon propagator.

Now, just below eq. (7.76), the authors say that in a scattering process with non-zero $q^2$, we get a quantity,
$$frac{-ig_{munu}}{q^2}left(frac{e_0^2}{1-Pi(q^2)}right){underset{mathrm{mathcal{O}(alpha)}}{=}}frac{-ig_{munu}}{q^2}left(frac{e^2}{1-[Pi_2(q^2)-Pi_2(0)]}right)$$
where $Pi_2(q^2)$ is the $mathcal{O}(alpha)$ value of $Pi(q^2)$.

My specific question is, why do we subtract $Pi_2(0)$ in the denominator of the RHS of the above expression? Shouldn’t we just replace $Pi(q^2)$ with $Pi_2(q^2)$?

Thanks in advance.

EDIT:
I think the answer lies a posteriori in the discussion leading up to eq. (7.90). They show that the $mathcal{O}(alpha)$ shift in the electric charge is given by,
$$Pi_2(0)approx-frac{2alpha}{3piepsilon}$$
which blows up as $epsilonrightarrow0$ ($epsilon=4-d$, $d$ being the number of spacetime dimensions). The argument is:

What can be observed is the $q^2$ dependence of the effective electric charge (7.77).

[(7.77) basically refers to the denominator I was talking about in my original question.]