Expanding $\frac{1}{1-z-z^2}$ to a power series.

Question

How would you expand the analytic function $$\frac{1}{1-z-z^2}$$ to a series of the form $$\sum_{k=0}^\infty a_k z^k \, \, ?$$

Answer 1

Write $1-z-z^2=(a-z)(z+b)$ and and using this, write the partial fraction of $$ \frac 1{1-z-z^2}=\frac 1{a+b}\left(\frac 1 {a-z} +\frac 1 {b+z} \right) $$

Answer 2

Hint. Consider $$(1-z-z^2)\left(\sum a_kz^k\right)=1$$ as a formal polynomial equation, where the RHS is the polynomial $1=1+0z+0z^2+\dotsb$. Multiplying out the LHS and equating with the RHS term by term, we find that $$\eqalign{a_0\,&=&\;1\cr -a_0+a_1\,&=&\;0\cr -a_0 - a_1 + a_2\,&=&\;0 \cr -a_1 - a_2 + a_3 \,&=&\;0\cr -a_2 - a_3 + a_4 \,&=&\;0 \cr \vdots\;\;\; &&\;\, \vdots\cr}$$ which we can rearrange to $$\eqalign{a_0\,&=&\;1\\a_1\,&=&\;a_0\\a_2\,&=&\;a_0 + a_1\\ a_3\,&=&\;a_1 + a_2\\ a_4\,&=&\;a_2 + a_3 \\ \vdots \;\;&&\;\;\vdots}$$ Does this sequence look familiar$\ldots$?