$\text{Res}((5z^{9}+z^{11})\text{cos}(\frac{1}{z^3}),0)$

Question

$\cdot $ Calculate : $\text{Res}((5z^{9}+z^{11})\text{cos}(\frac{1}{z^3}),0)$

First, the singularity at $z_0=0$ is irremovable because the lim of $\text{cos}(\frac{1}{z^3})$ doesn't exist.

So the only way to calculate $\text{Res}((5z^{9}+z^{11})\text{cos}(\frac{1}{z^3}),0)$ would be by the Laurent series, $$(5z^{9}+z^{11})\text{cos}(\frac{1}{z^3})=\Bigg(5z^{9}+z^{11}\Bigg)\Bigg(1-\frac{1}{2z^6}+\frac{1}{4!z^{12}}+\sum_{n=3}^{\infty}\frac{(-1)^n}{2n!}\bigg(\frac{1}{z^3}\bigg)^{2n}\Bigg)$$

But I don't see how that would be.

José Carlos Santos · Accepted Answer · 2022-04-18T20:44:05.233

1

The residue is $\frac1{24}$. Indeed, the residue of $5z^9\cos\left(\frac1{z^3}\right)$ at $0$ is $0$, since that's the coefficient of $\frac1z$ in the Laurent series of $5z^9\cos\left(\frac1{z^3}\right)$ centered at $0$, which is$$5z^9-\frac52z^3+\frac5{4!}\frac1{z^3}-\cdots$$whereas the coefficient of $\frac1z$ in the Laurent series of $z^{11}\cos\left(\frac1{z^3}\right)$ centered at $0$ is $\frac1{24}$, since this series is$$z^{11}-\frac12z^5+\frac1{24}\frac1z-\cdots$$

edited Apr 18 '22 at 20:44

answered Apr 18 '22 at 18:21

José Carlos Santos

427,504

I didn't know about that, so if $g$ is the sum of $2$ function and I can write them as Laurent expansions, the residue is the $a_{-1}$ term of both series ? What if they don't agree ? – 領域展開 Apr 18 '22 at 18:34
The residue is the sum of the residues of those two functions. Since one of the residues is $0$ and the other one is $\frac1{24}$, the residue of your function is $0+\frac1{24}=\frac1{24}$. – José Carlos Santos Apr 18 '22 at 18:36
Oh, now I get it, thanks! – 領域展開 Apr 18 '22 at 18:39
The $z^2$ term in the Laurent series of $z^9\cos\left(\frac{1}{z^3}\right)$ should be a $z^3$. – Enforce Apr 18 '22 at 19:07
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@Enforce I've edited my answer. Thank you. – José Carlos Santos Apr 18 '22 at 20:44

$\text{Res}((5z^{9}+z^{11})\text{cos}(\frac{1}{z^3}),0)$

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