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

I know (from a previous exercise) that:
$\lim_{z \to 0} \frac{\sin z - z}{z \sin z} = 0$
$f(z) = \left\{\begin{matrix} \frac{1}{\sin z} + \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} \text{ if } z \neq 0\\ 0 \text{ if } z = 0 \end{matrix}\right.$
Is analytic in $\{ z \in \mathbb{C} / | z | < 2 \}$
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}$to [itex](-1) \cdot$ the princial part of $\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z$, otherwise the limit when $z \ to 0$ wouldn't be zero, even more, I think it would diverge (It could be $\neq 0$ if the principal part coincides but the constant isn't the same).

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't want to ask him without being sure.

Last edited:

I know (from a previous exercise) that:
$\lim_{z \to 0} \frac{\sin z - z}{z \sin z} = 0$
$f(z) = \left\{\begin{matrix} \frac{1}{\sin z} + \frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} \text{ if } z \neq 0\\ 0 \text{ if } z = 0 \end{matrix}\right.$
Is analytic in $\{ z \in \mathbb{C} / | z | < 2 \}$
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}$to [itex](-1) \cdot$ the princial part of $\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z$, otherwise the limit when $z \ to 0$ wouldn't be zero, even more, I think it would diverge (It could be $\neq 0$ if the principal part coincides but the constant isn't the same).

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't want to ask him without being sure.

Try to learn how to use long division of polynomials. It comes up not too infrequently in Complex Analysis. You know:

$$\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$$

Ok, now just divide that into 1. I know that sounds simple and unfortunately we can's display what the long-division process looks like here. I'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' remember what they call the various pieces of long division but you'll get x^2/2-x^4/120 and so on. Now do it again. Eventually you'll run out of principal terms and start getting Taylor terms. Well, that's your principal part. Also, Mathematica does give the power series for Csc(z).

Edit: our subset of Latex does not support multicolumn which is needed to format nicely polynomial division. However, if you're interested, here's a link to see what it look's like if you already don't know. This one is similar to this problem:

http://tex.stackexchange.com/questi...-typesetting-of-polynomial-long-division?rq=1

Last edited: