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how to find the Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$.

$$\frac{z}{z^2+1} = \frac{1}{2}\left( \frac{1}{z-3 + (3 + i)} + \frac{1}{z-3 + (3 - i)}\right)$$

Given $|z-3| < \sqrt{10}$ or $|z-3| > \sqrt{10}$, I think we have two different expansions, how do I proceed? Do I need to consider both cases and thus have two answers?

hasExams
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  • Consider the first term (the other is similar). If $|z-3|<|3+i|$ then write it in the form $$\frac{1}{3+i} \frac{1}{1+\frac{z-3}{3+i}}$$ and expand as a geometric series. If $|z-3| > |3+i|$ then do this for $$\frac{1}{z-3} \frac{1}{1+\frac{3+i}{z-3}}.$$ – WimC May 01 '13 at 19:23
  • Are you sure that you are asked to find the series for $|z-3|>2$? – Mhenni Benghorbal May 02 '13 at 13:49
  • @MhenniBenghorbal yes ... that's what question asks for. if it had mentioned for $\sqrt 10 $ it would have been much clearer. I guess two different expansion for it. – hasExams May 02 '13 at 13:51

1 Answers1

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A related problem. First convert to the partial fraction form

$$ \frac{z}{z^2+1}=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}. $$

Then, we consider the first term, since the second is the same

$$ \frac{1}{2}\frac{1}{z-i}=\frac{1}{2}\frac{1}{(z-3)+(3-i)}= \frac{1}{2}\frac{1}{z-3}\frac{1}{1+\frac{3-i}{z-3}}$$

$$ = \frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k\,(3-i)^k}{(z-3)^{k+1}}\, $$

where

$$ \Big|\frac{3-i}{z-3}\Big| <1 \implies |z-3| > |3-i| \implies |z-3| > {\sqrt{10}}. $$

The same can be done with the other term which has the form

$$ = \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k(3+i)^k}{(z-3)^{k+1}}\quad |z-3|>\sqrt{10}. $$

So, we have

$$ \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k(3-i)^k}{(z-3)^{k+1}}+\frac{1}{2}\frac{1}{z-3}\sum_{k=0}^{\infty} \frac{(-1)^k(3+i)^k}{(z-3)^{k+1}} $$

$$ =\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k((3-i)^k+(3+i)^k)}{(z-3)^{k+1}}\,,\quad |z-3|> \sqrt{10}. $$

Community
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Mhenni Benghorbal
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  • but isn't it valid for $|z-3|< \sqrt{3^2 + 1^2}$, don't we need to consider case $2< |z-3| < \sqrt{10}$ – hasExams Apr 30 '13 at 07:54
  • I mean we would also get the analytic part when we consider that region ... while for $|z-3| > \sqrt{10}$ we have only principal part. – hasExams Apr 30 '13 at 08:05
  • @Crocodile: The above derived series do not converge for $|z-3|<\sqrt{10}$. However, you can derive the series that converge for $|z-3|<\sqrt{10}$ – Mhenni Benghorbal Apr 30 '13 at 10:22
  • Did you use ratio test here? $\Big|\frac{3-i}{(z-3)^2}\Big| <1 $ but shouldn't this be $\Big|\frac{3-i}{(z-3)}\Big|< 1$ ?? Also I think the problem is with the geometric expansion of $\frac{1}{1 + \frac{3+i}{z-3}}$. to expand it geometrically we get $|z-3|> |3+i|$ – hasExams Apr 30 '13 at 13:12
  • I just downvoted because the answer is wrong and because, despite numerous visits to the site since @Crocodile's comments were posted, the OP failed to address the mathematical problem aptly described in said comments. – Did May 01 '13 at 08:32
  • @Did: Thanks for the comment. By the way, the whole answer is wrong! – Mhenni Benghorbal May 01 '13 at 10:30
  • @Crocodile: I made a correction. Off course, these expansions of geometric series type. However, I do not see for the moment how to find the Laurent series valid for $|z-3|>2$. I'll be looking more closely to the problem once time allows. – Mhenni Benghorbal May 01 '13 at 10:34