(This question stems from problem 3.3 of Elvang's and Huang's "Scattering Amplitudes in Gauge Theory and Gravity" book).
Consider the Parke-Taylor amplitude given as \begin{equation} A_n[1^- 2^- 3^+\cdots n^+] = \frac{\langle 12\rangle^4}{\langle 12\rangle\langle 23\rangle\cdots \langle n1\rangle}\tag{3.9} \end{equation} Now lets consider a $[1^-,2^-\rangle$ shift (in literature usually seen as $[i,j\rangle$) where the rules for the shift-vectors are then given as \begin{equation} |\hat{1}\rangle = |1\rangle, |\hat{2}\rangle = |2\rangle - z|1\rangle \end{equation} (where I don't care about the square-bracket versions). Then, the products we need to worry about are $\langle 12\rangle$ which under the transformation remains the same (since $\langle 11\rangle = 0$), and the other shifted products are $\langle 12\rangle$, $\langle 23\rangle$, and $\langle n1\rangle$. The shift for $\langle 23\rangle$ is \begin{equation} \langle\hat{2}\hat{3}\rangle = \langle 23\rangle - z\langle 13\rangle = \langle 23\rangle\left(1 - z\frac{\langle 13\rangle}{\langle 23\rangle}\right) \end{equation} while $\langle \hat{n}\hat{1}\rangle$ remains unchanged. I can then write the shifted-amplitude as \begin{equation} A_n[\hat{1}^-\hat{2}^-3^+\cdots n^+] = A_n[1^-2^-3^+\cdots n^+]\frac{1}{1-z\langle13\rangle/\langle 23\rangle} \end{equation} displaying the $\sim 1/z$ shift. We then have a simple pole when $$1 - z\langle 13\rangle/\langle 23\rangle = 0\implies z_{pole} = \langle 23\rangle/\langle 13\rangle.$$ Now I want to calculate the shifted amplitude at the residue of $A_n(z)/z$ at this pole and this is where I get confused since I am missing the point or even the final result since it seems that I would then have a denominator $$\sim \frac{1}{1-z\langle13\rangle/\langle 23\rangle}\frac{1}{z}$$ and then two possible poles $$z_{poles} = \{0,\langle 23\rangle/\langle 13\rangle\}.$$ So, I guess I am really asking is what should I compute exactly? The pole at $z=0$ which just gives back an overall factor, while the pole at the other one gives back the same. I could absolutely be missing something important when it comes to evaluating the residues on page 52 of the textbook, but I am not sure what. I did notice that if I take my answer in the $z\rightarrow\infty$ limit then I recover negative of $A_n(z)/z$ taken at the residue $z = \langle 13\rangle/\langle 23\rangle$ which I believe is suppose to show that the $B_n$ factor cancels exactly and then the Parke-Taylor amplitude is considered "valid." Any thoughts, clarifications, suggestions, or even approvals are appreciated.
