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How might one go about computing the residue of $\frac {z^2 + 3z - 1}{z+2}$? I understand it has a pole at -2 and that we should then expand the numerator in powers of 2, but the book seems to do it by inspection. How does it look when done methodically?

EDIT: I should clarify - I understand I can just plug -2 into the numerator but I do not know why.

user73041
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1 Answers1

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Here is how check the order of the pole and notice that for the case of a simple pole it equals the residue. Now, $z=-2$ is a simple pole then the residue is the coeffiecient of $a_{-1}$ and it is given by

$$ \lim_{z\to -2}(z+2)\frac {(z^2 + 3z - 1)}{(z+2)}=\dots\,. $$

Other related problems (I), (II).

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Mhenni Benghorbal
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  • I see, so it works because in this case the pole is simple. Is the $(z+2)$ out front because of the definition of the residue or? – user73041 Apr 25 '13 at 21:51
  • @user73041: Yes, this is the formula (see the link) for computing it in the case of a simple pole. – Mhenni Benghorbal Apr 25 '13 at 21:54
  • You can write $\dfrac{z^2+3z-1}{z+2}=az+b+\dfrac{c}{z+2}$, and $c$ is the so-called residue. Multiply both sides by $(z+2)$... But yes, the formula used above works for every simple pole! – Philippe Malot Apr 25 '13 at 21:55
  • Thank you both. Is it perhaps helpful to think as polynomials as terms in an expansion in general when dealing with simple poles then? – user73041 Apr 25 '13 at 21:56
  • Another appoach to find residues is by deriving the Laurent series, but some times it requires a lot of work. – Mhenni Benghorbal Apr 25 '13 at 21:59