Question

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

Answer 1

\(\frac{1}{z^{2}+z+1}\) about \(z_{0}\) = \(1+i\)

Answer 2

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

Answer 3

is it possible to factor \(z^2+z+1\) and mimic the work for generating function of fibonacci series ?

Answer 4

perhaps this might help
\[z^2+z+1 = \dfrac{z^3-1}{z-1}\]

Answer 5

\[\dfrac{1}{z^2+z+1}= \dfrac{1}{1-z^3} - \dfrac{z}{1-z^3}\]

Answer 6

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

Answer 7

*fibonacci sequence

Answer 8

https://d2m81yxg6b9itc.cloudfront.net/flex-sequence-1/processed/fibonacci-formula.60d0194eb1ccc3a725abd73435bc877e/full/540p/index.mp4

Answer 9

you may skip to 5th minute

Answer 10

Alright, ill watch that first then see what I can come up with.

Answer 11

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

Answer 12

\[\dfrac{1}{z^2+z+1}= \dfrac{1}{(z-a)(z-b)}=\dfrac{1}{b-a}\left[\dfrac{1}{a-z} - \dfrac{1}{b-z}\right]\]

Answer 13

ofcourse hoping to use the geometric series
\[\dfrac{1}{1-z} = \sum\limits_{n=0}^{\infty}z^n\]

Answer 14

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

Answer 15

that looks neat! honestly i never did these before, im just lifting my knowledge about "real" to "complex"

Answer 16

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 ^_^

Answer 17

Awesome!