Let $∑_{n=0}^∞c_n z^n $ be a representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ Let $∑_{n=0}^∞c_n z^n $ be a power series representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ and radius of convergence of the series.
Clearly this is a power series with center $z_0=0$, and $f(z)=\frac{1}{1-z-z^2 }$ is analytic, because it's represented by a power series. I also know that
$$c_n =\frac{1}{n!} f^{(n)}(0)$$
but this doesn't get me anywhere, I also try the special case of Taylor series, but nothing look like this. I wonder if any one would give  me a hint please.

Using the factorization $1 - z - z^2 = -(z + \varphi)(z + \bar{\varphi})$, where $\varphi = (1 + \sqrt{5})/2$ and $\bar{\varphi} = (1 - \sqrt{5})/2$, we write
\begin{align}\frac{1}{1 - z - z^2} &= \frac{1}{\sqrt{5}}\left(\frac{1}{z + \varphi} - \frac{1}{z + \bar{\varphi}}\right) \\
& = \frac{1}{\sqrt{5}}\left(\frac{1/\varphi}{(z/\varphi) + 1} - \frac{1/\bar{\varphi}}{(z/\bar{\varphi}) + 1}\right)\\
& = \frac{1}{\sqrt{5}}\left(\frac{-\bar{\varphi}}{1 - \bar{\varphi}z} + \frac{\varphi}{1 - \varphi z}\right)\\
& = \frac{1}{\sqrt{5}}\sum_{n = 0}^\infty [-\bar{\varphi}(\bar{\varphi} z)^n +  \varphi(\varphi z)^n]\\
& = \sum_{n = 0}^\infty \left(\frac{\varphi^{n+1} - \bar{\varphi}^{n+1}}{\sqrt{5}}\right)z^n.
\end{align}
So $c_n = (\varphi^{n+1} - \bar{\varphi}^{n+1})/\sqrt{5}$ (which is equal to $F_{n+1}$, the $(n+1)$st Fibonacci number) and the radius of convergence is $1/\varphi$.
