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The $\require{cancel}\operatorname{Res}\left( \frac{z^a}{(z^2 + 1)^2},i \right)$ = $\frac{(1-a)e^{a i \pi/2}}{4i}$ is supposed to be this where $a \neq 1$ and $-1 < a < 3$

But what I am getting is $$\frac{d}{dz}\frac{ \cancel{(z-i)^2} z^a e^{ai\pi/2}}{\cancel{(z-i)^2} (z +i)^2} = \frac{(z^{-1 + a} (-2 z + a (i + z)))}{(i + z)^3}e^{ai\pi/2}$$

if I evaluate at $z=i$, how does the $i^a$ term go away?

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  • $\begingroup$ Where does the $e^{ai\pi/2}$ term on the left hand side of the displayed equation come from? $\endgroup$ – Daniel Fischer Aug 14 at 20:36
  • $\begingroup$ So $i$ the pole lies on the angle $\theta = \pi/2$. Hence $z^a = exp(a \log z ) = \exp (a \log |z| + ai\theta)) \implies exp(a \log|z| + i a\pi/2)$ $\endgroup$ – Hawk Aug 14 at 20:41
  • $\begingroup$ But the $z^{a}$ is still there. $\endgroup$ – Daniel Fischer Aug 14 at 20:42
  • $\begingroup$ Yes because $\exp(a \log |z|) \exp(ai\pi/2) = |z|^a \exp(a i \pi/2)$ $\endgroup$ – Hawk Aug 14 at 20:43
  • $\begingroup$ But now you have something you cannot differentiate. And besides, you wrote $z^{a}e^{ai\pi/2}$, not $\lvert z\rvert^{a}e^{ai\pi/2}$. $\endgroup$ – Daniel Fischer Aug 14 at 20:45
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As $ord(f(z),i)=-2$ then if $g(z)=(z-i)^2f(z)$ the residue of $f$ in $i$ is given by $g'(i)$ where $f(z)=\frac{z^a}{(z^2+1)^2}$.

$g'(z)= \Big(\frac{z^a}{(z+i)^2}\Big)'=\frac{az^{a-1}(z+i)^2-z^a2(z+i)}{(z+i)^4}=\frac{z^{a-1}(z+i)(a(z+i)-2z)}{(z+i)^4}$

$g'(i)=i^a\frac{-i(1-a)}{4}$. Finally as $i^a=e^{ln(i^a)}=e^{aln(i)}=e^{a\frac{\pi i}{2}}$ and $-i=\frac{1}{i}$ it follows the answer given.

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We have $\frac{d}{dz}(\frac{z^{a}}{(z+i)^{2}})=\frac{az^{a-1}(z+i)^{2}-2z^{a}(z+i)}{(z+i)^{4}}=\frac{z^{a}(a\frac{(z+i)^{2}}{z}-2(z+i))}{(z+i)^{4}}$ so setting $z=i$ gives $\frac{4ai-4i}{8}i^{a}=\frac{-i(1-a)}{2}(e^{i\frac{\pi}{2}})^{a}=\frac{a-1}{2i}e^{ai\frac{\pi}{2}}$ since $i=e^{i\frac{\pi}{2}}$.

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