TITLE: Residue of high order pole
QUESTION [6 upvotes]: 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?

REPLY [16 votes]: 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.

REPLY [10 votes]: Expanding at $i$, let $w = z - i$:
$$\begin{array}{rcl}
\operatorname{Res}(f,z_0)
&=& \operatorname{Res}\left(\dfrac1{w^7 (w+2i)^7},0\right) \\
&=& \operatorname{Res}\left(\dfrac1{w^7}(w+2i)^{-7},0\right) \\
&=& \operatorname{Res}\left(\dfrac1{w^7}(2i+w)^{-7},0\right) \\
&=& (2i)^{-7} \operatorname{Res}\left(\dfrac1{w^7}(1-0.5iw)^{-7},0\right) \\
&=& (2i)^{-7} \displaystyle \operatorname{Res}\left(\dfrac1{w^7} \sum_{n=0}^\infty \binom {-7} n (-0.5iw)^n,0\right) \\
&=& \displaystyle (2i)^{-7} \binom {-7} 6 (-0.5i)^6 \\
&=& -\dfrac{231}{2048}i
\end{array}$$