anonymous
  • anonymous
Series expansion of 1/(z^2 + z + 1) about point 1+i
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\(\frac{1}{z^{2}+z+1}\) about \(z_{0}\) = \(1+i\)
anonymous
  • anonymous
I'm assuming there's a way to do this without painstakingly using the definition, but I'm not quite sure how. I'm perfectly fine getting the expansion about \(z_{0} = 0\), just not sure what to do different in order to center it about 1 + i
ganeshie8
  • ganeshie8
is it possible to factor \(z^2+z+1\) and mimic the work for generating function of fibonacci series ?

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ganeshie8
  • ganeshie8
perhaps this might help \[z^2+z+1 = \dfrac{z^3-1}{z-1}\]
ganeshie8
  • ganeshie8
\[\dfrac{1}{z^2+z+1}= \dfrac{1}{1-z^3} - \dfrac{z}{1-z^3}\]
anonymous
  • anonymous
That definitely makes it look a lot easier (still would have to try it). How does this mimic the work for generating the fibonacci series? I havent really messed with anythin fibonacci
ganeshie8
  • ganeshie8
*fibonacci sequence
ganeshie8
  • ganeshie8
https://d2m81yxg6b9itc.cloudfront.net/flex-sequence-1/processed/fibonacci-formula.60d0194eb1ccc3a725abd73435bc877e/full/540p/index.mp4
ganeshie8
  • ganeshie8
you may skip to 5th minute
anonymous
  • anonymous
Alright, ill watch that first then see what I can come up with.
ganeshie8
  • ganeshie8
\(\large e^{\pm i2\pi/3}\) are the roots of \(z^2+z+1\), call these roots \(a,b\) respectively. so we can factor \(z^2+z+1\) as \((z-a)(z-b)\)
ganeshie8
  • ganeshie8
\[\dfrac{1}{z^2+z+1}= \dfrac{1}{(z-a)(z-b)}=\dfrac{1}{b-a}\left[\dfrac{1}{a-z} - \dfrac{1}{b-z}\right]\]
ganeshie8
  • ganeshie8
ofcourse hoping to use the geometric series \[\dfrac{1}{1-z} = \sum\limits_{n=0}^{\infty}z^n\]
anonymous
  • anonymous
So from there, just do the appropriate manipulation? For example: \[\frac{ 1 }{ a-z } = \frac{ 1 }{ a-(1+i)-[z-(1+i)] } =\frac{ 1 }{ a-(1+i) }\cdot \frac{ 1 }{ 1-\frac{ z-(1+i) }{ a-(1+i) } }\]
ganeshie8
  • ganeshie8
that looks neat! honestly i never did these before, im just lifting my knowledge about "real" to "complex"
anonymous
  • anonymous
Yeah, these turn out pretty cool, just I didnt consider making the roots into an arbitrary a and b. If Id have done that itd have come out much easier than me trying to integrate to arctan(z) and then differentiate that series back, lol. But yeah, its interesting getting more into these taylor and laurent series expansions. Thanks for the help and the idea ^_^
ganeshie8
  • ganeshie8
Awesome!

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