Taylor expansion of $\frac{1}{2-z-z^2}$

Question

The problem is:

Find the Taylor expansion of $f(z):= \dfrac{1}{2-z-z^2}$ on the disc $|z| < 1$

So far I have used partial fractions to obtain $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{2+z}\right)$ which I then rewrite as $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{1-((-1)(1+z))}\right)$.

My problem is that we are told specifically to find the expansion on the disc $|z| < 1$, and whilst $\dfrac{1}{1-z}$ is valid for $|z| < 1$, $\dfrac{1}{1-((-1)(1+z))}$ is valid for $|1+z| < 1$. I am a guessing that we cannot just say $f(z) = \dfrac{1}{3}\left(\displaystyle\sum_{n=0}^\infty z^n + (-1)^n(1+z)^n\right)$ but am not sure as my notes don't have any examples of this sort of difficulty.

Answer 1

Hint: $\frac{1}{1-z}=1+z+z^2+\ldots$ as infinitely descending geometric progression. And $\frac{1}{2+z}=\frac{1}{2}\cdot\frac{1}{1-(-z/2)}=\ldots$.

Answer 2

$\frac{1}{2-z-z^2}=\frac{ 1 }{ 2} \frac{1}{1-\frac{1}{2}z-\frac{1}{2}z^2}=\frac {1}{2}\sum _{ n = 0 } ^ \infty(\frac { 1 } { 2 } (z+z^2))^{n}$.

Done.

Note that Since $|z|\lt 1 $, it follows that $|\frac {1}{2}(z + z ^ 2 )|\lt \frac {1}{2}(|z|+|z ^ 2|)\lt1$ so it is justified to use geometric series above.