# Finding Laurent series of this function

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

• 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)
– user208649
Commented Aug 11, 2020 at 18:00
• @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 $z=i$
– user790216
Commented Aug 11, 2020 at 18:04

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.

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