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I know how to find residuum simple function but now I have function

$$f(z)=\frac{z-\pi}{\sin^2z}$$

and I have to calculate residuum in $\pi$ (that is, $ \operatorname{Res}_{z=\pi}f(z)$).

When I calculate the limit in $\pi$ it's infinity. So in $\pi$ we have pole function. And I have problem with calculate times the pole functions. Can someone help me?

2 Answers 2

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One may note that it is a simple pole, hence

$$\text{Res}_{z=\pi}\frac{z-\pi}{\sin^2z}=\lim_{z\to\pi}\frac{(z-\pi)^2}{\sin^2z}=1$$

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    why it's order two? it should be not equal 0. If it will be than it's order two. But it's equal 0 – 2017-01-22
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    @zxc It's order two since $\sin^2z\sim z^2$, hence a pole of order 2. – 2017-01-22
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    but it's not $Res \frac{1}{sin^2z}$, it's $Res \frac{z-\pi}{sin^2z}$ Still i dont understand why it has order two. – 2017-01-22
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    @zxc Ah, I thought you had already tried taking the residue of $1/\sin^2z$ and that was your attempt. Sorry for my misinterpretation. – 2017-01-22
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    and when you calculate this limit it's equal 0 but it shouldn't equal 0. for example $\frac{sinz}{z}$ in $0$ has order one beacuse $\lim\limits_{z->0}\frac{d}{dz}\frac{sinz}{z}$ it's not equal 0 so it has order one – 2017-01-22
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    @zxc Sorry, I fixed it. – 2017-01-22
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    so the order pole in $\pi$ is one ? – 2017-01-22
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    @zxc Yes. And I think your definition of the order of a pole is wrong. For simplicity, let $z'=z-\pi$ to take advantage of the period of sine so your problem reduces to $$\text{Res}_{z'=0}\frac{z'}{\sin^2z'}$$ – 2017-01-22
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52246/discussion-between-zxc-and-simply-beautiful-art). – 2017-01-22
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We have $f(z)=\frac{z-\pi}{\sin^2z}=\frac{z-\pi}{(z-\pi)^2\{1-\frac{(z-\pi)^2}{4!}.8+\frac{(z-\pi)^4}{6!}.32-\cdots\}}=\frac{1}{(z-\pi)\{1-\frac{(z-\pi)^2}{4!}.8+\frac{(z-\pi)^4}{6!}.32-\cdots\}}$ .

Now $\operatorname{Res}_{z=\pi}f(z)=\displaystyle\lim_{z\to\pi}\frac{z-\pi}{(z-\pi)\{1-\frac{(z-\pi)^2}{4!}.8+\frac{(z-\pi)^4}{6!}.32-\cdots\}}=\displaystyle\lim_{z\to\pi}\frac{1}{\{1-\frac{(z-\pi)^2}{4!}.8+\frac{(z-\pi)^4}{6!}.32-\cdots\}}=1$