I am asked to find $\text{Res}(i)$ for $R(z)=\frac{2z+3}{(z-i)(z^2+1)}$. Noticeably, $R(z)=\frac{2z+3}{(z-i)^2(z+i)}$.
While I understand how to find the corresponding partial fraction decomposition for the above equation, I am confused about the nature of $\text{Res}(i)$. Should $\text{Res}(i)$ be the coefficient for the $\frac{A}{(z-i)^2}$ term? Or the coefficient for the $\frac{B}{(z-i)}$ term? Or somehow both?
Thank you for your help!
$z=i$ is a double pole of $R(z)=\frac{2z+3}{(z-i)^2(z+i)}$, hence $$\text{Res}\left(R(z),z=i\right) = \lim_{z\to i}\frac{d}{dz}\left((z-i)^2 R(z)\right)=\lim_{z\to i}\frac{d}{dz}\left(\frac{2z+3}{z+i}\right) $$ and $$\text{Res}\left(R(z),z=i\right) =\lim_{z\to i}\frac{2i-3}{(z+i)^2}=\color{red}{\frac{3-2i}{4}}.$$ We have $$ R(z) = \frac{A}{(z-i)^2}+\frac{\color{red}{B}}{z-i}+\frac{C}{z+i}+Q(z) $$ where $Q(z)$ is an entire function and the wanted residue is $B$.