Finding the fourth derivative in order to get residue seems me very complicated, is there another way? $$Res\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right) }^{ 5 } } \right) =\lim _{ z\rightarrow i }{ \frac { 1 }{ 4! } \frac { { d }^{ 4 } }{ { d }{ z }^{ 4 } } \left( { \left( z-i \right) }^{ 5 }\frac { { e }^{ iz } }{ { \left( z-i \right) }^{ 5 }{ \left( z+i \right) }^{ 5 } } \right) } $$
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You can try the general Leibniz rule – Jakobian Jul 11 '18 at 15:57
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What do you need this for? If you are only interested in the result, then have a look: https://www.wolframalpha.com/input/?i=residue+of++e%5E%7Bix%7D%2F(x%5E2%2B1)%5E5++at+x%3Di – klirk Jul 11 '18 at 15:58
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Don't know if this is helpful, but you could also try to integrate the function around the singularity. – klirk Jul 11 '18 at 15:59
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I need another way as i mentioned above,wolframalpha gives an answer which i don't want before getting solution – haqnatural Jul 11 '18 at 16:00
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@klirk,finding integral demands to find residue either – haqnatural Jul 11 '18 at 16:03
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Not necessarily. You could take a loop, parametrize it and calculate the integral without knowing anything about the residue theorem – klirk Jul 11 '18 at 16:11
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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 $e^{it}=\sum_{k=0}^\infty\frac{(it)^k}{k!}$.
Markus Scheuer
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