Finding the singular points and residues of a function --- help

Question

can anyone help?

$\ f(z)=\frac{\sin(z)+1}{\sin^2(z)}$

Find the singular points of this function, classify them, and compute residues in these points. Try to demonstrate different methods for the computation of residues.

What are the singular points of this function and what methods should I use to find the residues? Thanks

Answer 1

We have $\ f(z)=\frac{\sin z+1}{\sin^2z}$.

Each zero $z_0$ of $\sin$ is an isolated singularity of $f$. Since $z_0$ is a zero of order $2$ of $\sin^2$ and not a zero of $ \sin z+1$, f has a pole of order $2$ at $z_0$.

We have, with $g(z):=(z-z_0)^2f(z)$:

$Res(f; z_0)= \lim_{z \to z_0}g'(z)$.

Your turn: what are the zeros of $\sin$ ?

Answer 2

Hint: The roots of the denominator are all poles of order 2 Hence you have poles of order 2 for $-\pi ,0, \pi , 2\pi ,...$ . As for the residues , use the following formula: $$\operatorname{Res}(f,z_k) =(\lim_{z\to z_k}(z-z_k)^2f(z))^{'}$$. Another way to calculate the residue in the specific excercise is to find the Laurent expansion of these functions.

Answer 3

$$\newcommand{\Res}{\operatorname*{Res}} \begin{align} \Res_{z=k\pi}\left(\frac{\sin(z)+1}{\sin^2(z)}\right) &=\Res_{z=k\pi}\left(\frac1{\sin(z)}\right)+\Res_{z=k\pi}\left(\frac1{\sin^2(z)}\right)\\ &=\Res_{z=0}\left(\frac1{\sin(z+k\pi)}\right)+\Res_{z=0}\left(\frac1{\sin^2(z+k\pi)}\right)\\ &=(-1)^k\Res_{z=0}\left(\frac1{\sin(z)}\right)+\Res_{z=0}\left(\frac1{\sin^2(z)}\right)\\[3pt] &=(-1)^k\cdot1+0\\[12pt] &=(-1)^k \end{align} $$ Since $\lim\limits_{z\to0}\frac{z}{\sin(z)}=1$ and $\frac1{\sin^2(z)}$ is an even function.