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I need to find the Laurent series and find the region of convergence for: \begin{equation} \frac{2z+i}{z(z+i)} \end{equation} About point $z=i$, which I've split into partial fractions to get: \begin{equation} \frac{1}{z}+\frac{1}{z+i} \end{equation} But I'm unsure on how to apply the definition of a Laurent series to obtain it for this equation at $z=i$.

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  • $\begingroup$ Since those two don't have poles at $i$, it is the same as the taylor-maclaurin series at $i$ (for that reason I am wondering if you might mean at $z=-i$ if this is an exercise that's suppose to get you comfortable with the pole part of a Laurent series) $\endgroup$
    – user208649
    Commented Aug 11, 2020 at 18:00
  • $\begingroup$ @TokenToucan I think this is just basic Laurent series practice (it's a follow on question from a MacLaurin series question), the question definitely says about point $z=i$ $\endgroup$
    – user790216
    Commented Aug 11, 2020 at 18:04

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You should rewrite both of the terms such that they can be expressed as a geometric series. For $\frac{1}{z+\mathrm i}$ this would be

$$\frac{1}{z+\mathrm i}=\frac{1}{2\mathrm i +(z-\mathrm i)}=\frac{1}{2\mathrm i}\frac{1}{1-\frac{\mathrm i}{2}(z-\mathrm i)}.$$

Then use the known formula

$$\sum_{k=0}^\infty q^k=\frac{1}{1-q}$$

for $\vert q\vert<1$. The other term works similarly.

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    $\begingroup$ I don't think your first expression is correct, although the idea is right. $\endgroup$
    – user208649
    Commented Aug 11, 2020 at 18:01
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    $\begingroup$ The OP ask for series around $i$ $\endgroup$ Commented Aug 11, 2020 at 18:02
  • $\begingroup$ Corrected both issues, thanks! $\endgroup$ Commented Aug 11, 2020 at 18:10
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    $\begingroup$ @Vercassivelaunos Thanks, I understand all the top steps, when using the known formula, do I just substitute $q$ for $\frac{i}{2}(z-i)$? $\endgroup$
    – user790216
    Commented Aug 11, 2020 at 18:14
  • $\begingroup$ @GroupTheory14: Exactly. And from the condition $\vert q\vert<1$ you can also gain information about the region of convergence. $\endgroup$ Commented Aug 11, 2020 at 18:20

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