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anonymous
 one year ago
Series expansion of 1/(z^2 + z + 1) about point
1+i
anonymous
 one year ago
Series expansion of 1/(z^2 + z + 1) about point 1+i

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\frac{1}{z^{2}+z+1}\) about \(z_{0}\) = \(1+i\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.2is it possible to factor \(z^2+z+1\) and mimic the work for generating function of fibonacci series ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2perhaps this might help \[z^2+z+1 = \dfrac{z^31}{z1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\dfrac{1}{z^2+z+1}= \dfrac{1}{1z^3}  \dfrac{z}{1z^3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That 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
 one year ago
Best ResponseYou've already chosen the best response.2you may skip to 5th minute

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, ill watch that first then see what I can come up with.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\large e^{\pm i2\pi/3}\) are the roots of \(z^2+z+1\), call these roots \(a,b\) respectively. so we can factor \(z^2+z+1\) as \((za)(zb)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\dfrac{1}{z^2+z+1}= \dfrac{1}{(za)(zb)}=\dfrac{1}{ba}\left[\dfrac{1}{az}  \dfrac{1}{bz}\right]\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2ofcourse hoping to use the geometric series \[\dfrac{1}{1z} = \sum\limits_{n=0}^{\infty}z^n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So from there, just do the appropriate manipulation? For example: \[\frac{ 1 }{ az } = \frac{ 1 }{ a(1+i)[z(1+i)] } =\frac{ 1 }{ a(1+i) }\cdot \frac{ 1 }{ 1\frac{ z(1+i) }{ a(1+i) } }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that looks neat! honestly i never did these before, im just lifting my knowledge about "real" to "complex"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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 ^_^
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