Finding the residue Finding the fourth derivative in order to get residue seems me  very complicated, is there another way? $$Res\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right)  }^{ 5 } }  \right) =\lim _{ z\rightarrow i }{ \frac { 1 }{ 4! } \frac { { d }^{ 4 } }{ { d }{ z }^{ 4 } } \left( { \left( z-i \right)  }^{ 5 }\frac { { e }^{ iz } }{ { \left( z-i \right)  }^{ 5 }{ \left( z+i \right)  }^{ 5 } }  \right)  } $$
 A: Since the residuum  is the coefficient of $(z-i)^{-1}$ in the Laurent-series expansion of $\frac{e^{iz}}{(z^2+1)^5}$ at  $z=i$, we can  expand the function and extract  the coefficient.

We  obtain
  \begin{align*}
\color{blue}{\mathrm{Res}}&\color{blue}{\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right)  }^{ 5 } }  \right)}\\
&=\mathrm{Res}\left(t=0,\frac{e^{i(t+i)}}{t^5(t+2i)^5}\right)\tag{1}\\
&=[t^{-1}]\frac{e^{i(t+i)}}{t^5(t+2i)^5}\tag{2}\\
&=[t^4]\frac{e^{i(t+i)}}{(t+2i)^5}\tag{3}\\
&=\frac{1}{(2i)^5}[t^4]e^{i(t+i)}\sum_{j=0}^\infty\binom{-5}{j}\left(\frac{t}{2i}\right)^j\tag{4}\\
&=\frac{1}{32ie}[t^4]e^{it}\sum_{j=0}^{\infty}\binom{j+4}{j}\left(-\frac{t}{2i}\right)^j\tag{5}\\
&=\frac{1}{32ie}\left(\frac{i^0}{0!}\binom{8}{4}\left(-\frac{1}{2i}\right)^4+\frac{i^1}{1!}\binom{7}{3}\left(-\frac{1}{2i}\right)^3
+\frac{i^2}{2!}\binom{6}{2}\left(-\frac{1}{2i}\right)^2\right.\\
&\qquad\qquad\left.+\frac{i^3}{3!}\binom{5}{3}\left(-\frac{1}{2i}\right)^1+\frac{i^4}{4!}\binom{4}{4}\left(-\frac{1}{2i}\right)^0\right)\tag{6}\\
&=\frac{1}{32ie}\left(\frac{35}{8}+\frac{35}{8}+\frac{15}{8}+\frac{5}{12}+\frac{1}{24}\right)\\
&\,\,\color{blue}{=-\frac{133i}{384e}}
\end{align*} 

Comment:


*

*In (1) we shift the residuum to $0$ by setting $t=z-i$.

*In (2) we use the coefficient of $[t^n]$ operator to denote the coefficient of $t^n$ in a series.

*In (3) we apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$.

*In (4) we factor out $(2i)^5$ and use the binomial series expansion.

*In (5) we use the binomial identity $\binom{-p}{q}=\binom{p}{q}(-1)^q$.

*In (6) we select the coefficient of $t^4$, by recalling $e^{it}=\sum_{k=0}^\infty\frac{(it)^k}{k!}$.
