- #1

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I know (from a previous exercise) that:

[itex]\lim_{z \to 0} \frac{\sin z - z}{z \sin z} = 0[/itex]

I had to prove that:

[itex]

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.

[/itex]

Is analytic in [itex]\{ z \in \mathbb{C} / | z | < 2 \}[/itex]

From this result I have to deduce that the principal part of Laurent's expansion of [itex]\frac{1}{\sin z} [/itex] is like

[itex]-\frac{1}{z}-2 \displaystyle \sum_{n=1}^{\infty} \frac{\pi^{2n}}{z^{2n+1}}[/itex]

I suppose they meant at [itex]z_{0} = 0[/itex], but they didn't state it anywhere (I suspect something in the exercise isn't correct).

What I did was:

[itex]\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}}[/itex]

And both parts are analytics.

Then I tried to use that:

[itex]\frac{1}{\sin z} = f(z)[/itex]

[itex]\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} = g(z)[/itex]

Then

[itex]\displaystyle \int_{C} (f(z)+g(z))z^{n} dz = 0 \forall n \in \mathbb{N}[/itex] (Because it's analytic).

Then we have

[itex]\displaystyle \int_{C} f(z)z^{n} dz = - \displaystyle \int_{C} g(z)z^{n} dz [/itex]

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}[itex] to [itex](-1) \cdot[/itex] the princial part of [itex]\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z[/itex], otherwise the limit when [itex]z \ to 0[/itex] wouldn't be zero, even more, I think it would diverge (It could be [itex] \neq 0[/itex] if the principal part coincides but the constant isn't the same).

I think the person who made the exercise did a mistake [itex]-\frac{1}{z}-2 \frac{1}{z} \displaystyle \sum_{n=1}^{\infty} \underbrace{\frac{\pi^{2n}}{z^{2n}}}_{\text{here}}[/itex] inverting the fraction, but I may be wrong and I don't want to ask him without being sure.

[itex]\lim_{z \to 0} \frac{\sin z - z}{z \sin z} = 0[/itex]

I had to prove that:

[itex]

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.

[/itex]

Is analytic in [itex]\{ z \in \mathbb{C} / | z | < 2 \}[/itex]

From this result I have to deduce that the principal part of Laurent's expansion of [itex]\frac{1}{\sin z} [/itex] is like

[itex]-\frac{1}{z}-2 \displaystyle \sum_{n=1}^{\infty} \frac{\pi^{2n}}{z^{2n+1}}[/itex]

I suppose they meant at [itex]z_{0} = 0[/itex], but they didn't state it anywhere (I suspect something in the exercise isn't correct).

What I did was:

[itex]\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}}[/itex]

And both parts are analytics.

Then I tried to use that:

[itex]\frac{1}{\sin z} = f(z)[/itex]

[itex]\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z} = g(z)[/itex]

Then

[itex]\displaystyle \int_{C} (f(z)+g(z))z^{n} dz = 0 \forall n \in \mathbb{N}[/itex] (Because it's analytic).

Then we have

[itex]\displaystyle \int_{C} f(z)z^{n} dz = - \displaystyle \int_{C} g(z)z^{n} dz [/itex]

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}[itex] to [itex](-1) \cdot[/itex] the princial part of [itex]\frac{z^{2}+\pi^{2}}{z^{3}-\pi^{2}z[/itex], otherwise the limit when [itex]z \ to 0[/itex] wouldn't be zero, even more, I think it would diverge (It could be [itex] \neq 0[/itex] if the principal part coincides but the constant isn't the same).

I think the person who made the exercise did a mistake [itex]-\frac{1}{z}-2 \frac{1}{z} \displaystyle \sum_{n=1}^{\infty} \underbrace{\frac{\pi^{2n}}{z^{2n}}}_{\text{here}}[/itex] inverting the fraction, but I may be wrong and I don't want to ask him without being sure.

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