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I want to show that he taylor series of $f(z)=\frac{1}{1-z+z^2}$ at zero is: $$\sum_{n}^{\infty}a_nz^n$$

where $a_0=a_1=0$, $a_2=0$ and $a_{n+3}=-a_n$. what is the radius of convergence ?

My attempt;(by the way I would much rather hints or explanations rather than full answers if possible)

First Using Taylor's theorem we have that $a_n=\frac{1}{2\pi i}\int\frac{f(z)}{(z-z_0)^{n+1}}dz$

Then by Cauchy's integral formula for derivatives we can say $a_n=\frac{1}{2\pi i}\int\frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{f^n(z_0)}{n!}$ and using this neat little formula we can compute the coefficients

$a_0=f(0)=1$

$a_1=f'(0)=1$

$a_2=f''(0)/2=0/2=0$

{Personal Question 1: how could one go about proving $a_{n+3}=-a_n$, I thought to do something like $a_{n+3}=\frac{f^{n+3}(0)}{(n+3)!}=\frac{1}{(n+3)!}\frac{d^n(f^3(0)}{dz^n}$ But this comes out as zero.}

Next I calculated the radius of convergence by the ration test $R=\lim_{n \to\infty}|\frac{a_{n+1}}{a_n}|=f'(0)/n+1=0$

And finially in attempting to show that the taylor series representation is given by $\sum_{n}^{\infty}a_nz^n$,I used a Taylor expansion to show $f(z)=f(0)+f'(0)z+\frac{f''(0)z^2}{2}+...+\frac{f^n(z)}{n!}$

Then I used the formula found from combining the Taylor and Cauchy theorems as above where we see that $a_n=\frac{f^n(z_0)}{n!}$. which means $f(z)=\sum a_nz^n$

Personal question number 2: is it cheating to use a Taylor expansion as i did here or do you think this is what the question was looking for ?

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$$\frac{1}{1-z+z^2}=\frac{1+z}{1+z^3}=\frac{(1+z)(1-z^3)}{1-z^6} = (1+z-z^3-z^4)\sum_{k\geq 0} z^{6k} $$ is enough and it clearly shows that the radius of convergence is $1$.

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\begin{align}\frac1{1-z+z^2}&=\frac{1+z}{1+z^3}\\&=(1+z)(1-z^3+z^6-z^9+\cdots)\text{ (if $|z|<1$)}\\&=1+z-z^3-z^4+z^6+z^7-z^9-z^{10}+\cdots\end{align} Besides, note that if you wish to express this as $\displaystyle\sum_{n=0}^\infty a_nz^n$, then the sequence of coefficients is defined by$$a_n=\begin{cases}1&\text{ if }n=0\vee n=1\\0&\text{ if }n=2\\-a_{n-3}&\text{ otherwise.}\end{cases}$$

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