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In my textbook, I have this example:

$$f(z) = \frac{9z+i}{z^3+z}.$$ Find the residue.

So the book says it has a simple pole at $i$ but doesn't it also have a simple pole at $-i$?

I can see the denominator is $z(z^2 + 1) = z(z+i)(z-i)$ so doesn't it have three simple poles: $z = 0, z = i, z = -i$?

The residue that the book gives is:

$$\operatorname{Res}_{z=i} \frac{9z+i}{z(z^2+1)} = \lim_{z \to i} (z-i) \frac{9z+i}{z(z+i)(z-i)} = \frac{9z+i}{z(z+i)}\bigg|_{z=i} = \frac{10i}{-2} = -5i$$

But aren't there other poles?

Jwan622
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  • How did you set up your contour? $-i$ may not be in it – imranfat Nov 29 '20 at 00:38
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    Yes, there are other poles, but we don't have to consider them in order to compute the residue at $z=i$ – saulspatz Nov 29 '20 at 00:38
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    If the example really asks "find the residue," then it is ambiguous; one should ask "find the residue at [singularity] $i$" or something like that. – angryavian Nov 29 '20 at 00:40
  • My book does not say anything of the sort. – Jwan622 Nov 29 '20 at 23:00
  • Is the problem $\frac{9z+1}{z^3+z}$ or $\frac{9z+i}{z^3+z}$? You have both formulae in your question. – Math Keeps Me Busy Nov 30 '20 at 17:26
  • It's an i in the numerator @MathKeepsMeBusy – Jwan622 Nov 30 '20 at 18:29
  • thanks for fixing it. – Math Keeps Me Busy Nov 30 '20 at 18:31
  • In total agreement with @angryavian , I would remind you that any text book is not written by a divinity but by a human being. Text have plenty of mistakes, especially in the exercises and their answers. – Lubin Nov 30 '20 at 18:55
  • @angryavian I think my book was a bit ambiguous but I may have misinterpreted. IT says that f(z) has a simple pole and i, but not that it was the only simple pole. I think I misinterpreted and just wanted to make sure that it clearly had other simple poles. – Jwan622 Nov 30 '20 at 20:02

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