Laurent series of $f(z)=\text{cosec}(z)$ that is valid at $\frac{1}{2}$

Question

I am trying to find a Laurent series for $f(z)=\text{cosec}(z)$ up to the $z^3$ term about $z=0$ that is valid at $\frac{1}{2}$.

Now $f$ has singularities at $z=k\pi, \ \ k\in\mathbb{Z}$. So if we want a series that's valid at $\frac{1}{2}$, I think the region we're interested in is $$0<|z|<\pi.$$

For the Laurent series, I have tried using the Taylor expansion of $\sin(z)$, but I don't know how to proceed.

A hint would be very helpful.

Answer 1

Since $\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots,$ $0$ is a simple pole of $\csc$. Besides, $\csc$ is an odd function. So, around $0$,$$\csc(z)=\frac{a_{-1}}z+a_1z+a_3z^3+\cdots=\frac{a_{-1}+a_1z^2+a_3z^4+\cdots}z.$$Furthermore, $\sin(z)\csc(z)=1$, which means that$$\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots\right)\left(a_{-1}+a_1z^2+a_3z^4+\cdots\right)=1.$$In particular, $a_{-1}=1$, $-\frac{a_{-1}}{3!}+a_1=0$, and $a_3-\frac{a_1}{3!}+\frac{a_{-1}}{5!}=0$, which means that $a_1=\frac16$ and that $a_3=\frac7{360}$. So$$\csc(z)=\frac1z+\frac z6+\frac{7z^3}{360}+\cdots$$And, yes, it converges to $\csc z$ whenever $0<\lvert z\rvert<\pi$.

Answer 2

Hint:

Rewrite as $$\operatorname{cosec}z=\frac1{\sin z}=\frac1{z-\dfrac{z^3}6+\dfrac{z^5}{120}+o(z^5)}= \frac1z\,\frac1{1-\dfrac{z^2}6+\dfrac{z^4}{120}+o(z^4)}$$ and expand $\;\frac1{1-\tfrac{z^2}6+\tfrac{z^4}{120}+o(z^4)}$ up to order $3$, for instance with a division of $1$ by $\;1-\frac{z^2}6+\frac{z^4}{120}$ by increasing powers up to order 3.

Using this division, you obtain an equality: $$1=\Bigl(1+\frac {z^2}6+\frac{7z^4}{360}\Bigr)\Bigl(1-\frac{z^2}6+\frac{z^4}{120}\Bigr)+ z^4 r(z), $$ for some polynomial $r(z)$, so that $$\frac1{1-\tfrac{z^2}6+\tfrac{z^4}{120}+o(z^4)}=1+\frac {z^2}6+\frac{7z^4}{360}+o(z^4)$$ and dividing by $z$, we obtain $$\operatorname{cosec}z=\frac1z+\frac {z}6+\frac{7z^3}{360}+o(z^3).$$