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.
 A: Here is the one way which will work for any rational function with a denominator you can factor, and which has simple roots.  A modification of the procedure will also allow you to tackle the case of repeated roots:

*

*Use partial fraction decomposition.

*If you have a linear term $az+b$ in the denominator, rewrite it as $a(z-i+i)+b = a(z-i)+ai+b = (ai+b)[1 - \frac{-a}{ai+b}(z-i)]$.

*Each rational functions is now written in a form where you should be able to recognize the sum of a geometric series.

*Combine the two geometric series.

*To find the radius of convergence, either apply a classic test like the Ratio test, or just use the fact that holomorphic functions always have disks of convergence which are "as large as possible".  So the radius of convergence will be the distance of the closest pole to $z=i$.

Thomas mentions an easier way for this particular function in the comments to the OP though :)
A: You can expand $$\frac{1}{(z-i+i)^2+z-i+i+1}=\frac{1}{(z-i)^2+(2i+1)(z-i)+i}$$ as
$$\frac{1}{(z-i)^2+(2i+1)(z-i)+i}=\sum_{\alpha,\beta}\binom{-1}{\alpha,\beta,-1-\alpha-\beta}(z-i)^{2\alpha+\beta}(2i+1)^{\beta}i^{-1-\alpha-\beta}$$ where $$\binom{-1}{\alpha,\beta,-1-\alpha-\beta}=\frac{1(-1)(-2)(-3)\cdots}{\alpha!\beta!(-1-\alpha-\beta)(-2-\alpha-\beta)(-3-\alpha-\beta)\cdots}$$ and the products continue, cancelling out when appropriate. For example, let $\alpha=5,\beta=2.$ Then
$$\binom{-1}{5,2,-8}=\frac{(-1)(-2)(-3)(-4)(-5)(-6)(-7)\color{red}{(-8)(-9)(-10)\cdots}}{5!2!\color{red}{(-8)(-9)(-10)\cdots}}=\frac{(-1)^77!}{5!2!}=-21$$ The whole thing converges when $$|((z-i)^2+(2i+1)(z-i))|<1,$$ or in the region defined by the implicit equation
$$2 x + 3 x^2 + 2 x^3 + x^4 - 2 y - 4 x y - y^2 + 2 x y^2 + 2 x^2 y^2 + y^4<3$$
A: $f$ can be decomposed as: $\displaystyle f(z)=\frac1{i\sqrt3}\Bigl( \frac1{z-j}-\frac1{z-\bar{\jmath}}\Bigr)$ with $j=\exp\bigl( i\frac{2\pi}3 \bigr)$, thus $\displaystyle f^{(n)}(z)=\frac{(-1)^n\, n!}{i\sqrt3}\Bigl( \frac1{(z-j)^{n+1}}-\frac1{(z-\bar{\jmath})^{n+1}}\Bigr).$
Since we are going to replace $z$ by $i$, we note first that
$\begin{cases}
i-j=\frac12+(1-\frac{\sqrt3}2)i=r_1\exp(i\frac{\pi}{12}),& \text{where}\ r_1=\sqrt{2-\sqrt3}\in\mathopen]0,1\mathclose[,\\[4pt]
i-\bar{\jmath}=\frac12+(1+\frac{\sqrt3}2)i=r_2\exp(i\frac{5\pi}{12}),& \text{where}\ r_2=\sqrt{2+\sqrt3}=\frac{1}{r_1}>1.
\end{cases}$
Now we get
$\begin{eqnarray}
\frac{f^{(n)}(i)}{n!}&=&\frac{(-1)^n}{i\sqrt3}\Bigl[ r_2^{n+1}\exp\bigl(-i(n+1)\frac{\pi}{12}\bigl) - r_1^{n+1}\exp\bigl(-i(n+1)\frac{5\pi}{12}\bigr) \Bigr]\\
&=&\frac{(-1)^n}{i\sqrt3}\Bigl[ r_2^{n+1}\cos\bigl((n+1)\frac{\pi}{12}\bigr)-r_1^{n+1}\cos\bigl((n+1)\frac{5\pi}{12}\bigr)\\
&&\qquad{}+i\bigl\{ r_1^{n+1}\sin\bigl((n+1)\frac{5\pi}{12}\bigr)-r_2^{n+1}\sin\bigl((n+1)\frac{\pi}{12}\bigr)\bigr\}
\Bigr]
\end{eqnarray}$
which doesn't seem to have a significantly more “compact” form.
As for the radius of convergence, the precedent formula provides
$\displaystyle
\left|\frac{f^{(n)}(i)}{n!}\right|^2
=\frac13 \Bigl[ r_1^{2n+2}+r_2^{2n+2}-2\cos\bigl((n+1)\frac{\pi}{3} \bigr) \Bigr]\sim\frac{r_2^{2(n+2)}}3$ as $n\to\infty$,
which implies that $\lim\limits_{n\to\infty}\left|\frac{f^{(n)}(i)}{n!}\right|^{1/n}=r_2$.
The radius is then $\frac1{r_2}=r_1$.
We can observe that the numbers $x_k=(2-\sqrt3)^k+(2+\sqrt3)^k$ appearing above are integers.
They satisfy the relations: $x_0=2$, $x_1=4$ and $x_{k+2}=4x_{k+1}-x_k$ ($k\geqslant2$).
A: $$\frac{1}{z^2+z+1}=\sum_{n=0}^\infty a_n (z-i)^n$$
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 $z^2+(1+2i)z+i$
