So I am having issues with the following function:
$f(z)=\frac{1}{\sin(z)}-\frac{1}{z}$
I need to find that $a_{-1}$ in the Laurent series of $\frac{1} {\sin (z)}$ at $0$.
So firstly I went and proved that $f(z)=\frac{1}{\sin(z)}-\frac{1}{z}$ has a removable singularity at $0$.
It should be clear that $\frac{1}{z}$ has a simple pole at $z = 0$.
We also know the Taylor series of $\sin(z)$ and indeed the roots of $\sin(z)$; namely $n\pi,n \in \mathbb Z$.
Thus ord$(\sin(z),n\pi) = 1$ so that ord${(\frac{1}{\sin(z)} ,n\pi)}= −1$;
i.e. $\frac{1}{\sin(z)}$ has simple poles at $n\pi$. Thus $f$ has simple poles at $n\pi$ for $n \in \mathbb Z/(0)$.
What happens at $z = 0$? Rewriting gives
$f(z) = \frac{z−\sin(z)}{ z \sin(z)}$
where
$z−\sin(z) =\frac{z^3}{ 3!}- \frac{z^5}{ 5!}+ …$
and
$z \sin(z) = z^2−\frac{z^4}{4!}…$
so that
ord$(z−\sin(z),0) = 3$, and ord$(z\sin(z),0) = 2$
ord$(f,0) = 1$.
Thus $f$ has a removable singularity at $z = 0$. Defining $f(0) = a_0 = 0$ makes $f$ differentiable at $z = 0.$
How do I use what I have shown to find $a_{-1}$ in the Laurent Series of $\frac{1}{\sin(z)}$
Best Answer
Since $f$ has a removable singularity at $0$ and since $f(0)=0$ we can write
$$f(z)=a_1z+a_2z^2+a_3z^3+....$$
in a neighborhood of $0$.
Since $\frac{1}{ \sin z}=f(z)+\frac{1}{z}$ we have
$$\frac{1}{ \sin z}=\frac{1}{z}+a_1z+a_2z^2+a_3z^3+....$$
in a deleted neighborhood of $0$.
Thus $a_{-1}=1.$
A furthr possibility:$g(z):=\frac{1}{ \sin z}$ has a simple pole at $0$. Thus
$$a_{-1}=Res(g;0)= \lim_{z \to 0}zg(z)=1.$$