$\newcommand{\de}{\delta}
\newcommand{\ga}{\gamma}
\newcommand{\si}{\sigma}
\newcommand{\ep}{\epsilon}$
Take any real $n>0$ and any $z:=\zeta\in(0,\infty)$. Let $\ep\downarrow0$. Note that for $\ep>0$ 
\begin{equation}
	\left|(t-i\ep)^{-n}-(t+i\ep)^{-n}\right|=2(t^2+\ep^2)^{-n/2}|\sin(nu)|
	=2\ep^{-n}\frac{|\sin(nu)|}{|\sin u|^n},  
\end{equation}
where $u:=\arctan(\ep/t)$, so that $t=\ep\cot u$. So,  
 
\begin{equation}
	I_n(z,\ep)=4\ep^{1-n}J_n(\de),  
\end{equation}
where  
\begin{equation}
	J_n(\de):=\int_\de^{\pi/2}\frac{|\sin(nu)|}{(\sin u)^{n+2}}\,du. 
\end{equation}
and 
\begin{equation}
	\de:=\arctan\frac\ep z\downarrow0
\end{equation}

Let $c=c(\ep)\downarrow0$ in such a manner that $c^{-n-1}=o(\ep^{-n})$; e.g., we may take $c=\ep^{n/(n+2)}$. Then 
\begin{equation}
	0\le J_n(c)\ll\int_c^{\pi/2}\frac1{u^{n+2}}\,du\ll c^{-n-1}=o(\ep^{-n}).  
\end{equation}
On the other hand, $\de\asymp\ep=o(c)$ and 
\begin{equation}
	J_n(\de)-J_n(c)=\int_\de^c\frac{|\sin(nu)|}{(\sin u)^{n+2}}\,du
	\sim\int_\de^c\frac{nu}{u^{n+2}}\,du
	=\de^{-n}-c^{-n}\sim\de^{-n}\sim z^n\ep^{-n}. 
\end{equation}
So, 
\begin{equation}
	J_n(\de)=J_n(\de)-J_n(c)+J_n(c)\sim z^n\ep^{-n}
\end{equation}
and 
\begin{equation}
	I_n(z,\ep)=4\ep^{1-n}J_n(\de)\sim4z^n\ep^{1-2n}.   
\end{equation}