The corrected propagator is given by $$\Delta'(q)=\frac{1}{q^2+m^2-\Pi^*(q^2)-i\epsilon}$$
($\Pi^*$ is the sum of all irreducible one-particle amplitudes) I get that the residue of the original propagator around the pole $q^2=-m^2$ is $$\frac{1}{2\pi i}\oint_{\text{around }q^2=-m^2} \frac{dq^2}{q^2+m^2-i\epsilon}=\lim_{q^2\rightarrow -m^2}\frac{q^2+m^2}{q^2+m^2}=1$$ and that the corrected propagator must have the same residue $$\frac{1}{2\pi i}\oint \Delta'(q)dq^2=1$$ So how does the condition $$\left[\frac{d\Pi^*(q^2)}{dq^2}\right]_{q^2=-m^2}=0$$ ensure the second integral above?

EDIT: Devouring complex analysis literature. Have already edited some things that weren't quite right. For anyone interested, I'm using Weinberg Vol 1 and this in section 10.3, ~p. 430.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7