# Residue calculation check.

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?

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

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.
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 $$i=e^{i\frac{\pi}{2}}$$.