Field Strength Renorm in Peskin&Schroeder On page 237 in PS we have (the unnumbered equation after eq. 7.58)
$$\mathcal{P} \sim \frac{iZ}{p^2-m^2-iZ\,\mathrm{Im}M^2(p^2)}$$
but after deriving it myself I obtained
$$\mathcal{P} \sim \frac{iZ}{p^2-m^2-iZ\,\mathrm{Im}M^2(p^2)-iZ\frac{\mathrm{d}\,\mathrm{Im}\, M^2}{\mathrm{d}\,p^2}\cdot(p^2-m^2)+\dots}$$
why do they omit the derivative term? Why is it considered small?
Note: My mistake was that I also expanded the imaginary part of $M^2$...please see answer below for solution. 
 A: You are certainly correct that there are other terms in the sum. However, the derivative term is zero by the renormalization conditions for the scalar field and the other terms are assumed to be small (when $p^2 $ is far from $m^2$ the diagrams as small anyways). For completeness here the derivation:
By performing an infinite sum over $ 1PI $ diagrams we can write the propagator as
\begin{equation} \tag1
\frac{ i }{ p ^2 - m _0 ^2 + M ^2  ( p ^2 )}
\end{equation} 
where $ - i M ^2 $ is the value of all $ 1PI $ contributions. 
The problem is that due to the optical theorem if a particle can decay it means it will have an imaginary component to $ M ^2 $ so we can no longer identify the position of the pole with the mass. Instead we define a particle's mass by:
\begin{equation} 
m ^2 - m _0 ^2 - \mbox{Re} ( M ^2 ( m ^2 ) ) = 0 
\end{equation} 
Using the equation above we write the propagator in terms of physical quantities as:
\begin{align} \tag2
\frac{ i }{ p ^2 - m ^2 - M ^2 ( p ^2 ) } & \approx \frac{ i Z }{ p ^2 - m ^2  + Z\mbox{Re} M ^2 ( m ^2 ) - ZM ^2 ( p ^2 ) }
\end{align} 
At this point as you said we expand the correction:
\begin{equation} 
\mbox{Re} M ^2 ( p ^2 ) = \mbox{Re} M ^2 ( m ^2 ) + \frac{ d \mbox{Re} M ^2  }{ d p ^2 } \bigg|_{ p ^2 = m ^2 } \left( p ^2 - m ^2 \right) + ...
\end{equation} 
The first order term cancels in the propagator and the second term is zero due to renormalization conditions of the scalar field (see equation 10.28 in Peskin). Thus around the pole we can approximately write:
\begin{align} 
\frac{ i }{ p ^2 - m ^2 - M ^2 ( p ^2 ) } & \approx \frac{ i Z }{ p ^2 - m ^2   - iZ\mbox{Im} M ( p ^2 )  }
\end{align} 
A: Starting from 
\begin{equation} \tag1
\mathcal{P}\sim\frac{ \mathrm{i} }{ p ^2 - m _0 ^2 + M ^2  ( p ^2 )}
\end{equation}
and as Jeff points out, by the Optical theorem*, $M^2$ (for a particle that decays) can have a nonzero imaginary part. Hence one defines the physical mass $m$ of the particle, not through 
\begin{equation} 
m ^2 - m _0 ^2 - M ^2 ( m ^2 ) = 0, 
\end{equation}
as one usually does, but rather through 
\begin{equation} \tag2
m ^2 - m _0 ^2 - \mbox{Re}\,  M ^2 ( m ^2 ) = 0. 
\end{equation}
Now $M^2(p^2)\tag3 = \mathrm{Re}\,M^2(p^2)+\mathrm{i}\cdot\mathrm{Im}\,M^2(p^2).$
Expanding the real part only 
\begin{equation} \tag4
\mbox{Re}\, M ^2 ( p ^2 ) = \mbox{Re}\, M ^2 ( m ^2 ) + \frac{ \mathrm{d}\, \mbox{Re}\, M ^2  }{ \mathrm{d} \,p ^2 } \bigg|_{ p ^2 = m ^2 } \left( p ^2 - m ^2 \right) + ...
\end{equation}
and plugging ($2$), ($3$) and ($4$) into our original propagator ($1$) we find 
\begin{equation} 
\mathcal{P}\sim\frac{ \mathrm{i} }{ (p ^2 - m^2)\underbrace{\left(1-\frac{ \mathrm{d}\, \mbox{Re}\, M ^2  }{ \mathrm{d} \,p ^2 }\bigg|_{ p ^2 = m ^2 }\right)}_{Z^{-1}} -\mathrm{i}\cdot\mathrm{Im}\, M ^2  ( p ^2 )+\mathcal{O}\left((p ^2 - m^2)^2\right)}
\\=
\frac{ \mathrm{i}Z }{ (p ^2 - m^2) -\mathrm{i}Z\cdot\mathrm{Im}\, M ^2  ( p ^2 )+\mathcal{O}\left((p ^2 - m^2)^2\right)}
\end{equation}
which is what we wanted to show. 
\begin{equation}
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\end{equation}
*(See equation $(7.49)$ PS)
