Unitarity and amplitudes In Bootstrap and Amplitudes: A Hike in the Landscape of Quantum Field Theory there are few statements about analytical structure of amplitudes.

I want to understand statement:

In a local theory of massless scalars, they can have simple poles and the residue of such a pole is, by
unitarity, a product of lower-point amplitudes.

I understand this property from Feynman rules for tree level amplitude, but I didn't get role of unitarity.
I wanna understand, how unitarity related to such property? How residues will change, if we lose unitarity?
 A: Unitarity ($SS^\dagger = 1$) dictates that
\begin{equation}\label{key}
 T-T^\dagger = iTT^\dagger
\end{equation}
For tree-level $2\rightarrow 2$ scattering, we have then that
$$
\langle p_1,p_2|T|p_3,p_4\rangle - \langle p_1,p_2|T|p_3,p_4\rangle^* = i\langle p_1,p_2|TT^\dagger|p_3,p_4\rangle.
$$
Using $\langle p_1,p_2|T|p_3,p_4\rangle = (2\pi)^4\delta^{(4)}(p_1 + p_2 - p_3 - p_4)\mathcal{A}[12\rightarrow 34]$, and inserting a complete set of one particle states (to stay at tree-level), we can write this as
\begin{align}
 2\text{Im}(\mathcal{A}[12\rightarrow 34]) &= 2\pi\sum_k\int \frac{d^3k}{2E_k}\delta^{(4)}(p_1+p_2 - k)\mathcal{A}[12\rightarrow k]\mathcal{A}^*[k\rightarrow 34]\\
 &= 2\pi\sum_k\int d^4k\delta(k^2)\delta^{(4)}(p_1+p_2 - k)\mathcal{A}[12\rightarrow k]\mathcal{A}^*[k\rightarrow 34]
\end{align}
Now, the left hand side is the imaginary part of the 4pt amplitude, which we will take to have numerators $n_i$ and propagators $p_i^2 + i\epsilon$, where $i$ labels the ways we can arrange a particle exchange (the $s,t,u$ channels). Thus, we have
\begin{equation}
 2\text{Im}\left(\sum_k\frac{n_k}{k^2+i\epsilon} + \text{contact}\right) = 2\pi\sum_k\int d^4k\delta(k^2)\delta^{(4)}(p_1+p_2 - k)\mathcal{A}[12\rightarrow k]\mathcal{A}^*[k\rightarrow 34],
\end{equation}
In a local theory of massless scalars, we can write the imaginary part of the propagator as
$$
\text{Im}\left(\frac{1}{p^2 + i\epsilon}\right) = \frac{1}{2i}\left(\frac{1}{p^2 + i\epsilon} - \frac{1}{p^2 - i\epsilon}\right) = \frac{-\epsilon}{p^4 + \epsilon^2}.
$$
This seems like it vanishes for $\epsilon \rightarrow 0$, which, by the optical theorem, means that your amplitude must be zero for real external momenta. However, this is misleading and only true when the propagator is off-shell. Recognising the fact that the last term above is the nascent dirac delta function, we learn that
$$
\lim_{\epsilon\rightarrow 0}\frac{-\epsilon}{p^4 + \epsilon^2} = \pi\delta(p^2).
$$
Plugging this in, we find that, as the propagator goes on shell, we have
\begin{equation}
 2\pi\sum_kn_k\delta(k^2) = 2\pi\sum_k\int d^4k\delta(k^2)\delta^{(4)}(p_1+p_2 - k)\mathcal{A}[12\rightarrow k]\mathcal{A}^*[k\rightarrow 34].
\end{equation}
Or, in other words, the numerators of the tree-level amplitudes factorize into two lower-point amplitudes (the residues) as the propagator goes on-shell
\begin{equation}
 n_k = \int d^4k\delta^{(4)}(p_1+p_2 - k)\mathcal{A}[12\rightarrow k]\mathcal{A}^*[k\rightarrow 34]
\end{equation}
We note now a major problem: the right hand side is actually zero due to momentum conservation, and this is probably the reason most books don't discuss the optical theorem at tree-level. This is due to the fact that Lorentz-invariant three-particle amplitudes vanish on-shell by virtue of the fact that $p_i\cdot p_j = 0$ for all $i,j$ due to momentum conservation. However, this is not true if we use spinor helicity variables and assume complex momentum, which is exactly what the bootstrapping amplitudes program does.
