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I was practising for an exam and I had some trouble with the following excersice:

$$f(z)= \frac{1}{z \sin z}$$

a. Find the pole and its order.

$$\frac{1}{z(z-z^3/3!+ z^5/5! + \cdots)}= \frac{1}{z^2(1-z^2/3!+ z^4/5! + \cdots)}$$

So the pole $z=0$ has order 2, but what about the other poles? $n\pi$?

b. Find the residue in this pole of $f$.

$$\lim_{z \to 0} \frac{\mathrm d}{\mathrm dz} \frac{z}{\sin z}= \lim_{z \to 0} \frac{\sin z-z \cos z}{\sin^2z}$$ and now I don't know how to continue.

Thanks in advance!

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  • $\begingroup$ Please consider using \sin, \cos and \lim to get $\sin$, $\cos$ and $\lim$ instead of $sin$, $cos$ and $lim$. Moreover, please use $$ at the starts and $$ at the end to get code displayed on its own line. Finally, please consider using \cdots to get $\cdots$ instead of ... to get $...$ If you want dots at the bottom then use \ldots to get $\ldots$ $\endgroup$ – Fly by Night Jan 24 '14 at 20:01
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Jan 24 '14 at 20:39
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You had

$$\frac1{z^2\left(1-\frac{z^3}6+\mathcal O(z^4)\right)}=\frac1{z^2}\left(1+\frac{z^3}6+\frac{z^6}{36}+\ldots\right)=\frac1{z^2}+\frac z6+\ldots$$

so the residue is zero.

Also, using what you did and applying l'Hospital's rule:

$$\lim_{z\to 0}\frac{\sin z-z\cos z}{z^2}\stackrel{\text{l'H}}=\lim_{z\to 0}\frac{z\sin z}{2z}=0$$

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  • $\begingroup$ Thank you for your reply! I don't really get the first equality. Can you explain it to me? Moreover why did you calculated the above limit? I thought I had to calculate: $$\lim_{z \to 0} \frac{\sin z-z \cos z}{\sin^2z}$$ But L'Hospital gives also zero for this. $\endgroup$ – Leslie Jan 26 '14 at 15:04
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DonAntonio answered all but your second question for part (a), so I'll discuss that.

To find the order of the poles at $z = n \pi$, compute $$ \lim_{z \to n \pi} \frac{\sin(z)}{z - n \pi} \, . $$ If the limit is nonzero, that means it's a simple pole. If the limit is zero, compute $$ \lim_{z \to n \pi} \frac{\sin(z)}{(z - n \pi)^k} $$ for larger values of $k$ until you get a nonzero limit.

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  • $\begingroup$ Thank you SpamIAm, I don't really get how to continue, $$\lim_{z^\to n\pi} \frac{\sin(z)}{z-n\pi} = cos(n\pi)$$ and this is 1 when n is even and -1 when n is odd. So the limit does'nt exist? And is it also possible to solve this exercise with $$\lim_{z\to n\pi} \frac{z-n\pi}{z\sin(z)}$$ and see if this limit exists? (if it does the pole has order 1 otherwise I will increase the power of $$(z-n\pi)^k$$) But how do I evaluate this limit? $\endgroup$ – Leslie Jan 26 '14 at 14:54
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    $\begingroup$ Oo I think I understand it. The limit does exist but it is different for n even or odd! $\endgroup$ – Leslie Jan 26 '14 at 15:11

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