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I'm trying to compute the residue $\displaystyle\operatorname{Res}\left(\frac{1}{(z^2+1)^7},i\right)$.
I know that there is the formula:
$$\operatorname{Res}(f,z_0)=\frac{1}{(m-1)!}\lim_{z\rightarrow z_0 }[(z-z_0)^mf(z)]^{(m-1)}$$
for a pole with order $m$.
But I'm pretty sure that I should not try to compute the 6th derivative of $\dfrac{1}{(z+i)^7}$.
Is there another way to compute the residue beside this formula?
Hmm. \begin{align*} \lim_{z\to i}\left[\frac{1}{(z+i)^7}\right]^{(6)}&= \lim_{z\to i}\left[(z+i)^{-7}\right]^{(6)} \\ &=\lim_{z\to i}\left[-7(z+i)^{-8}\right]^{(5)} \\ &=\lim_{z\to i}\left[(-7)(-8)(z+i)^{-9}\right]^{(4)} \dots \end{align*} Calculating the derivatives doesn't seem too bad.
Expanding at $i$, let $w = z - i$:
$$\begin{array}{rcl} \operatorname{Res}(f,z_0) &=& \operatorname{Res}\left(\dfrac1{w^7 (w+2i)^7},0\right) \\ &=& \operatorname{Res}\left(\dfrac1{w^7}(w+2i)^{-7},0\right) \\ &=& \operatorname{Res}\left(\dfrac1{w^7}(2i+w)^{-7},0\right) \\ &=& (2i)^{-7} \operatorname{Res}\left(\dfrac1{w^7}(1-0.5iw)^{-7},0\right) \\ &=& (2i)^{-7} \displaystyle \operatorname{Res}\left(\dfrac1{w^7} \sum_{n=0}^\infty \binom {-7} n (-0.5iw)^n,0\right) \\ &=& \displaystyle (2i)^{-7} \binom {-7} 6 (-0.5i)^6 \\ &=& -\dfrac{231}{2048}i \end{array}$$
