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anonymous

  • one year ago

Series expansion of 1/(z^2 + z + 1) about point 1+i

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  1. anonymous
    • one year ago
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    \(\frac{1}{z^{2}+z+1}\) about \(z_{0}\) = \(1+i\)

  2. anonymous
    • one year ago
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    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

  3. ganeshie8
    • one year ago
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    is it possible to factor \(z^2+z+1\) and mimic the work for generating function of fibonacci series ?

  4. ganeshie8
    • one year ago
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    perhaps this might help \[z^2+z+1 = \dfrac{z^3-1}{z-1}\]

  5. ganeshie8
    • one year ago
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    \[\dfrac{1}{z^2+z+1}= \dfrac{1}{1-z^3} - \dfrac{z}{1-z^3}\]

  6. anonymous
    • one year ago
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    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

  7. ganeshie8
    • one year ago
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    *fibonacci sequence

  8. ganeshie8
    • one year ago
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    you may skip to 5th minute

  9. anonymous
    • one year ago
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    Alright, ill watch that first then see what I can come up with.

  10. ganeshie8
    • one year ago
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    \(\large e^{\pm i2\pi/3}\) are the roots of \(z^2+z+1\), call these roots \(a,b\) respectively. so we can factor \(z^2+z+1\) as \((z-a)(z-b)\)

  11. ganeshie8
    • one year ago
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    \[\dfrac{1}{z^2+z+1}= \dfrac{1}{(z-a)(z-b)}=\dfrac{1}{b-a}\left[\dfrac{1}{a-z} - \dfrac{1}{b-z}\right]\]

  12. ganeshie8
    • one year ago
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    ofcourse hoping to use the geometric series \[\dfrac{1}{1-z} = \sum\limits_{n=0}^{\infty}z^n\]

  13. anonymous
    • one year ago
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    So from there, just do the appropriate manipulation? For example: \[\frac{ 1 }{ a-z } = \frac{ 1 }{ a-(1+i)-[z-(1+i)] } =\frac{ 1 }{ a-(1+i) }\cdot \frac{ 1 }{ 1-\frac{ z-(1+i) }{ a-(1+i) } }\]

  14. ganeshie8
    • one year ago
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    that looks neat! honestly i never did these before, im just lifting my knowledge about "real" to "complex"

  15. anonymous
    • one year ago
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    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 ^_^

  16. ganeshie8
    • one year ago
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    Awesome!

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