Find the principal part of 1/sin(z) at z0=0

[SqueeSpleen]

I know (from a previous exercise) that:
\lim_{z \to 0} \frac{\sin z - z}{z \sin z} = 0
I had to prove that:
<br /> f(z) = \left\{\begin{matrix}<br /> \frac{1}{\sin z} + \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} \text{ if } z \neq 0\\ <br /> 0 \text{ if } z = 0<br /> \end{matrix}\right.<br />
Is analytic in \{ z \in \mathbb{C} / | z | &lt; 2 \}
From this result I have to deduce that the principal part of Laurent's expansion of \frac{1}{\sin z} is like
-\frac{1}{z}-2 \displaystyle \sum_{n=1}^{\infty} \frac{\pi^{2n}}{z^{2n+1}}
I suppose they meant at z_{0} = 0, but they didn't state it anywhere (I suspect something in the exercise isn't correct).
What I did was:
\frac{1}{\sin z} + \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} = \underbrace{(-\frac{1}{\sin z} + \frac{1}{z})}_{= 0 \text{ when } z \to 0} + \displaystyle \sum_{n=1}^{\infty} \frac{z^{2n-1}}{\pi^{2n}}
And both parts are analytics.

Then I tried to use that:
\frac{1}{\sin z} = f(z)
\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} = g(z)
Then
\displaystyle \int_{C} (f(z)+g(z))z^{n} dz = 0 \forall n \in \mathbb{N} (Because it's analytic).
Then we have
\displaystyle \int_{C} f(z)z^{n} dz = - \displaystyle \int_{C} g(z)z^{n} dz
I did the something with negative powers (Now I think I made my mistake here, I'll think it later).

Anyway, not even wolframalpha gives me the expansion I'm supposed to find.

All I can guess but I'm not sure how to justify it, is that the principal part of [/itex]\frac{1}{\sin z}to (-1) \cdot the princial part of \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z, otherwise the limit when z \ to 0 wouldn&#039;t be zero, even more, I think it would diverge (It could be \neq 0 if the principal part coincides but the constant isn&#039;t the same).<br /> <br /> I think the person who made the exercise did a mistake -\frac{1}{z}-2 \frac{1}{z} \displaystyle \sum_{n=1}^{\infty} \underbrace{\frac{\pi^{2n}}{z^{2n}}}_{\text{here}} inverting the fraction, but I may be wrong and I don&#039;t want to ask him without being sure.

 

[jackmell]

I know (from a previous exercise) that:
\lim_{z \to 0} \frac{\sin z - z}{z \sin z} = 0
I had to prove that:
<br /> f(z) = \left\{\begin{matrix}<br /> \frac{1}{\sin z} + \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} \text{ if } z \neq 0\\ <br /> 0 \text{ if } z = 0<br /> \end{matrix}\right.<br />
Is analytic in \{ z \in \mathbb{C} / | z | &lt; 2 \}
From this result I have to deduce that the principal part of Laurent's expansion of \frac{1}{\sin z} is like
-\frac{1}{z}-2 \displaystyle \sum_{n=1}^{\infty} \frac{\pi^{2n}}{z^{2n+1}}
I suppose they meant at z_{0} = 0, but they didn't state it anywhere (I suspect something in the exercise isn't correct).
What I did was:
\frac{1}{\sin z} + \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} = \underbrace{(-\frac{1}{\sin z} + \frac{1}{z})}_{= 0 \text{ when } z \to 0} + \displaystyle \sum_{n=1}^{\infty} \frac{z^{2n-1}}{\pi^{2n}}
And both parts are analytics.

Then I tried to use that:
\frac{1}{\sin z} = f(z)
\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} = g(z)
Then
\displaystyle \int_{C} (f(z)+g(z))z^{n} dz = 0 \forall n \in \mathbb{N} (Because it's analytic).
Then we have
\displaystyle \int_{C} f(z)z^{n} dz = - \displaystyle \int_{C} g(z)z^{n} dz
I did the something with negative powers (Now I think I made my mistake here, I'll think it later).

Anyway, not even wolframalpha gives me the expansion I'm supposed to find.

All I can guess but I'm not sure how to justify it, is that the principal part of [/itex]\frac{1}{\sin z}to (-1) \cdot the princial part of \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z, otherwise the limit when z \ to 0 wouldn&#039;t be zero, even more, I think it would diverge (It could be \neq 0 if the principal part coincides but the constant isn&#039;t the same).<br /> <br /> I think the person who made the exercise did a mistake -\frac{1}{z}-2 \frac{1}{z} \displaystyle \sum_{n=1}^{\infty} \underbrace{\frac{\pi^{2n}}{z^{2n}}}_{\text{here}} inverting the fraction, but I may be wrong and I don&#039;t want to ask him without being sure.

<br /> <br /> Try to learn how to use long division of polynomials. It comes up not too infrequently in Complex Analysis. You know:<br /> <br /> \sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots<br /> <br /> Ok, now just divide that into 1. I know that sounds simple and unfortunately we can&#039;s display what the long-division process looks like here. I&#039;ll start it for you: z goes into 1 a total of 1/z right? Then 1/z times that expression is going to be 1-z^2/6+z^4/120+. . . Now subtract that from I dont&#039; remember what they call the various pieces of long division but you&#039;ll get x^2/2-x^4/120 and so on. Now do it again. Eventually you&#039;ll run out of principal terms and start getting Taylor terms. Well, that&#039;s your principal part. Also, Mathematica does give the power series for Csc(z).<br /> <br /> Edit: our subset of Latex does not support multicolumn which is needed to format nicely polynomial division. However, if you&#039;re interested, here&#039;s a link to see what it look&#039;s like if you already don&#039;t know. This one is similar to this problem:<br /> <br /> <a href="http://tex.stackexchange.com/questions/123731/replicate-a-typesetting-of-polynomial-long-division?rq=1" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://tex.stackexchange.com/questi...-typesetting-of-polynomial-long-division?rq=1</a>