I am trying to understand an example of where a series converges in Gamelin complex analysis textbook. The example and the part I don't understand are as follow.
Consider the Laurent series for $f(z) = (z^2-\pi^2)/sin(z)$ that is centered at 0 and that converges for $|z|=1$. What is the largest open set on which the series converges?
The part that I don't understand is the following. "Since $sin(z)$ has a simple zero at $\pi$, the function $sin(z)/(z-\pi)$ extends to be analytic and nonzero at $z = \pi$. Hence $(z^2-\pi^2)/sin(z)$ extends to be analytic at $z = \pi$". My question is why would the function $sin(z)/(z-\pi)$ eis analytic and nonzero at $ \pi$, wouldn't I have $0/0$ and that would not be defined? Also, why does this fact implies that $(z^2-\pi^2)/sin(z)$ is analytic at $z = \pi$? Thanks for your help!
The intuition is that when you consider this equation, $f(z) = \frac{(z^2 - \pi^2)}{sin z} $ This has a zero in both numerator and the denominator. If you consider the expansion of $ \sin z $ around $ z = \pi$ then you will see that there is a factor of $z -\pi$ in front of the series ( ie. no constant term)
Hence the zero from the denominator will cancel the zero from the numerator and f will be defined well and analytic at z = $\pi$.
Similarly with $z = -\pi$
The implication follows from the fact that if $g(z) \neq 0$ and is analytic, then $\frac{1}{g(z)}$ is as well.
Therefore the largest annulus would be $ 0 < |z| < 2 \pi $