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I am trying to find the co-efficient of $\frac{1}{z}$ in the expansion of $$\frac{(1+z^2)^{2n}}{z^{2n+1}}$$

I proceeded like this -

$$\frac{1}{z^{2n+1}}[1+(2n)z^2+\frac{(2n)(2n-1)z^4}{2!}+\dots+\frac{(2n)(2n-1)(2n-2)\dots(2n-n+1)z^{2n}}{n!}+\dots]$$

So, the required co-efficient is $$\frac{(2n)!}{(n!)^2}$$ However, according to the book the denominator has power $n$ not $2$. Where am i going wrong ?

user1176409
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Aman Mittal
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  • That could be a typo. Your approach and answer look correct to me. – Hans Engler Jun 18 '14 at 19:40
  • Can you add a bit of context? I can't really tell what it is about. Of course, the coefficient of $z^{-1}$ in $z^{-(2n+1)}(1+z^2)^{2n}$ is $\binom{2n}{n}$, as you computed. But what the book might expect instead, I cannot guess from what you posted. – Daniel Fischer Jun 18 '14 at 19:40
  • @DanielFischer I am trying to find the residue at $z=0$ for the said function. In the context of complex analysis. Part of this question i asked earlier http://math.stackexchange.com/questions/836445/use-calculus-of-residues-to-evlauate/836474 – Aman Mittal Jun 18 '14 at 19:43
  • Okay. And what precisely does the book say would be the result? – Daniel Fischer Jun 18 '14 at 19:45
  • @DanielFischer According to the book, the answer is $\frac{(2n)!\pi}{2^{2n-1}(n!)^n}$. I am able to get everything except the power of $n!$ in the denominator right. – Aman Mittal Jun 18 '14 at 19:47
  • That's a plain typo. – Daniel Fischer Jun 18 '14 at 19:53
  • @DanielFischer :) Thanks !! – Aman Mittal Jun 18 '14 at 19:54

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