Question is a fully solved exercise in Gamelin's complex analysis.
Exercise : Consider the Laurent series expansion for $\frac{1}{\sin z}$ that converges on the circle $\{|z|=4\}$. Find the coefficients $a_0,a_{-1},a_{-2},a_{-3}$. Determine the largest open set on which the series converges.
Solution : The only zeros of $\sin z$ are at the integral multiples of $\pi$. These are then the only singularities of $\frac{1}{\sin z}$, and they are all simple poles. The largest open annular set containing the cirlce and to which $\frac{1}{\sin z}$ extends analytically is then the annulus $\{\pi <|z|<2\pi\}$. The annulus is then the largest open set on which the Laurent series converges.
We have expansions $$\sin z =z-\cdots \rm{near~~}z=0$$ $$\sin z =-(z-\pi)-\cdots \rm{near~~}z=\pi$$ $$\sin z =-(z+\pi)-\cdots \rm{near~~}z=-\pi$$
Just by considering inverses, long divisions, we see that
$$\frac{1}{\sin z} =\frac{1}{z}+\rm{analytic}$$ $$\frac{1}{\sin z}=-\frac{1}{z-\pi}+\rm{analytic}$$ $$\frac{1}{\sin z} =-\frac{1}{z+\pi}+\rm{analytic}$$
We conclude that if $$f_1(z)=\frac{1}{z}-\frac{1}{z-\pi}-\frac{1}{z+\pi}$$ then $f_0(z)=1/\sin z-f_1(z)$ is analytic for $|z|>2\pi$. Thus, $$\frac{1}{\sin z}=f_0(z)+f_1(z)$$ is the Laurent decomposition of $1/\sin z$.
I do not understand what ever is written in bold. I see that $f(z)$ is analytic for $|z|>\pi$.. could not figure out why $f_0(z)$ is analytic for $|z|<2\pi$..
We have $$f_3(z)=\frac{3}{\sin z}-f_1(z)=\left(\frac{1}{\sin z}-\frac{1}{z}\right)+\left(\frac{1}{\sin z}-\frac{1}{z-pi}\right)+\left(\frac{1}{\sin z}-\frac{1}{z+\pi}\right)$$
Each function in the brackets is analytic at $0,\pi,-\pi$ respectively. does it imply sum is analytic?? That too in $|z|<2\pi$$??
$f_1(z)$ is just same as $f_3(z)$ except a constant.. so, if $f_3(z)$ is analytic then so would be $f_1(z)$..
Help me to fill the gaps..