<s>[INST] How to find the taylor series of $f(z)=\frac{1}{z^2+z+1}$ I'm trying to find the taylor expansion of the function $f(z)=\frac{1}{z^2+z+1}$ about $z_0=i$ and its radius of convergence. I'm not sure how to do it without differentiating.
 [/INST] $$\frac{1}{z^2+z+1}=\sum_{n=0}^\infty a_n (z-i)^n$$
Shift $z→z+i$
$$\frac{1}{z^2+(1+2i)z+i}=\sum_{n=0}^\infty a_n z^n$$
$$=\frac{1}{(z-r)(z-s)}=\frac{1}{r-s}\left(\frac{1}{z-r}-\frac{1}{z-s}\right)$$$$=\frac{1}{r-s}\left(\frac{1/s}{1-z/s}-\frac{1/r}{1-z/r}\right)$$$$=\frac{1}{r-s}\sum_{n=0}^\infty\left(\frac{1}{s^{n+1}}-\frac{1}{r^{n+1}}\right)z^n$$
Now that we found $a_n$, we plug it back in, or rather shift $z→z-i$
$$\frac{1}{z^2+z+1}=\frac{1}{r-s}\sum_{n=0}^\infty\left(\frac{1}{s^{n+1}}-\frac{1}{r^{n+1}}\right)z^n$$
$r,s$ are roots of $z^2+(1+2i)z+i$
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