0

Expanded to the Laurent Series at the deleted neighbourhood of $$z=i$$ and try to give the Convergence range

I'm trying to make $$\frac{1}{(1+z^2)^2}=\bigg(\frac{-1}{2z}\bigg)\bigg(\frac{1}{1+ z^2 }\bigg)\text {'}$$

$$\frac{1}{1+ z^2}={1-z^2+z^4-z^6...}=\sum_{n=0}^{\infty} {(-1)^ n(z^{2n})}$$

$$\bigg(\frac{1}{1+ z^2 }\bigg)\text {'}= \sum_{n=0}^{\infty} {(2n)(-1)^ n(z^{2n-1})}$$

so that

$$\frac{1}{(1+z^2)^2}=\bigg(\frac{-1}{2z}\bigg)\sum_{n=0}^{\infty} {(2n)(-1)^ n(z^{2n-1})}$$

$$=-\sum_{n=0}^{\infty} {(n)(-1)^n(z^{2n-2})}$$ But it seems not right...

awc789
  • 5
  • 5
  • Please learn use of mathjax for question asking: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Rhys Hughes May 03 '18 at 13:48
  • Are you trying to compute a Laurent series or an integral? It's hard to see what the first line has to do with the third line. – saulspatz May 03 '18 at 14:00
  • Do you mean $|z|>1$ or $|z|<1$? – Algebear May 03 '18 at 14:03
  • This doesn't seem to have anything to do with Laurent series. You haven't even told us the center--is it $i$ or $-i$? It looks more like you are trying to compute a Taylor series. Please clarify your question. – saulspatz May 03 '18 at 14:06
  • @RhysHughes Wow thank you，i am trying to learn it！ it's really useful ! – awc789 May 03 '18 at 14:18
  • @saulspatz i am so sorry, i have not made it clear, and now i learn to use mathjax, it's better – awc789 May 03 '18 at 15:08
  • @GuusPalmer sorry for say that, well, it's at the deleted neighbourhood of point &z=i$ – awc789 May 03 '18 at 15:09
  • @saulspatz sorry for that, i am just started to use Stack Exchange, it is at the deleted neighbourhood of point $z=i$ – awc789 May 03 '18 at 15:10
  • @awc789 Then you should have powers of $z-i$ shouldn't you? Everything you are doing is about $z=0$ – saulspatz May 03 '18 at 15:33

2 Answers2

0

Hint: Series for $\frac{1}{1+z^2}$ is $\frac{1}{1-(-z^2)}=\sum_{k=0}^{\infty}(-z^2)^k$.

Algebear
  • 1,674
0

I really can't follow what you are doing. You start by saying, "I'm trying to make $$\frac{1}{(1+z^2)^2}=\bigg(\frac{-1}{2z}\bigg)\bigg(\frac{1}{1+ z^2 }\bigg),$$ but you surely know that this isn't a true statement. I also don't see what it has to do with what follows.

Let me get you started: $$\frac{1}{(z^2+1)^2}=\frac{1}{(z+i)^2(z-i)^2} $$ Now $1/(z+i)^2$ is analytic at $z=i$ and so can be developed in a Taylor series about $z=i$. Once you do this, you can simply multiply the series by $1/(z-i)^2$ to get the desired Laurent series.

saulspatz
  • 53,131