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Determine the singular points of the following functions, the nature of these singular points and compute the residues in these points. $$(a)\:\dfrac{\cos z}{z^3},\qquad (b)\:\dfrac z{\sin z},\qquad(c)\:\dfrac{e^{z+10}}{z^{10}}.$$

Hi there - For $(a)$ and $(b)$, I know that the singularities are both $0$, with orders of $3$ and $10$ respectively.

I know that the singularity of $(b)$ is $\sin(z)=o$, i.e. $nĪ€$. But I am not sure of the order of this one. Would it not be infinity (the $n$'s can keep on increasing)?

Thanks for your help.

Hakim
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Archie Judd
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    Each singularity has it is own order. As for $\sin'(n\pi) \ne 0$ for each $n$, $\sin$ has first order zeros at $z = n\pi$, so $z/\sin z$ has first order poles at $z = n\pi$, $n \ne 0$, and a removable singularity at $z = 0$ (here the numerator is zero too and $z/\sin z \to 1$, $z \to 0$). – martini Aug 01 '14 at 11:53
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    Check this. – Mhenni Benghorbal Aug 01 '14 at 11:59

1 Answers1

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For (a) you can consider the power-series $cos(x)=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}$. Write down $\frac{cos(z)}{z^3}$ as a power-series and $a_{-1}$ is the residue in zero.

For (b) all singularities have the form $\pi n$. For $n=0$ the singulariy is removeable (L'hospital). For $n \neq 0$ consider the limit $\lim_{z \to n \pi} |\frac{z}{sin(z)}|$.You will obtain that the singularites are poles (Why?)

The order of the poles are then the first $k \in \mathbb N$ such that $\lim_{z \to n \pi} \frac{z(z-n\pi)}{sin(z)}$ exists. The value is also the residue in $n\pi$ (Why?)

(c) works similar

Marm
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