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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!

Esteban
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  • The phrase extends to be analytic means has an analytic extension, in the same way that has a continuous extension is used in elementary calculus and beginning real analysis courses. See, for example, my answer at What is a continuous extension?. As for why the resulting extended function is analytic, this should follow immediately from what your book has developed up to this point, especially with regard to simple simple zeros of analytic functions. – Dave L. Renfro Mar 23 '16 at 16:44
  • Thanks, Dave! I'll review the simple zeros of analytic functions section. – Esteban Mar 23 '16 at 20:34

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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 $

Aleks J
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  • When you say that the expansion of $sin z$ around $z = \pi$ has a factor of $\pi$ in front, how did you get that expansion? I mean, when I expand $sin z$ around $z = \pi$ I get $(z−\pi)−(z−\pi)^3/6+(z−\pi)^5/120−...$. Could you elaborate a bit more on this? – Esteban Mar 23 '16 at 20:44
  • Sorry, I meant a factor of $z - \pi$ see edited answer – Aleks J Mar 23 '16 at 22:36