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?
Best Answer
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.