I'm learning complex analysis, specifically (Laurent) series and residues, and need help with the following problem:
Construct a function $f(z)$ holomorphic in $\mathbb{C}\setminus\{i, -i\}$ with a pole of order $2$ at $z=i$ s.t. $\operatorname{Res}\left(f, i\right)=1$, a simple pole at $z=-i$ s.t. $\operatorname{Res}\left(f, -i\right)=2$ and a zero of order $3$ at $z=0$.
Here are my thoughts:
The function $1/(z-i)^2$ takes care of the requirement at $i$. And $1/(z+i)$ takes care of the requirement at $-i$. If I add them together, then the resulting function will have a pole of order $2$ at $i$ and a pole of order $1$ at $-i$. But how do I take care of the requirements for the residues?
Also we require a zero of order $3$ at $0$, i.e. a function of the form $z^3g(z)$, where $g(0)\neq0$.
I don't know how to combine all of the requirements together to get the desired function $f(z)$ (and how to take care of the residues).
A example of function that have a pole of order $p$ at $z_0$ and residue $r$ is given by $$f(z) = \frac{rz^{p-1}}{(z-z_0)^p}.$$ Indeed, to compute the residue, we can use the formula $$Res(f,z_0) = \lim_{z\to z_0}\frac{1}{(p-1)!}((z-z_0)^pf(z))^{(p-1)}=r.$$
Hence a example of function with pole of order $2$ with residue $1$ at $i$ and simple pole with residue $2$ at $-i$ is given by $$f(z) = \frac{z}{(z-i)^2}+\frac{2}{z+i}.$$ Now you must add $z^3$ in front of that but it will modify the residues. For exemple in the first term you will have to derive $z^4$ one time et evaluate at $i$. So you must change the coefficients to do the compensation. Finally (if no mistakes), you get $$f(z) = z^3 \left( \frac{iz}{4(z-i)^2}-\frac{2i}{z+i}\right).$$