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

1. anonymous

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

2. 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

3. ganeshie8

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

4. ganeshie8

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

5. ganeshie8

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

6. 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

7. ganeshie8

*fibonacci sequence

8. ganeshie8
9. ganeshie8

10. anonymous

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

11. 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 $$(z-a)(z-b)$$

12. ganeshie8

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

13. ganeshie8

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

14. anonymous

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

15. ganeshie8

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

16. 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 ^_^

17. ganeshie8

Awesome!