The question

For the following function,

i) Find the largest open disc about the given point $\displaystyle z_0$ in which f is analytic.

ii) find the first 3 non-zero terms of the Taylor series of f about $\displaystyle z_0$

iii) Find the coefficient $\displaystyle a_n$ of $\displaystyle (z - z_0)^n$ in the Taylor series of f about $\displaystyle z_0$ either as an explicit function of n, or give a recursive formula for $\displaystyle a_n$ as appropriate.

f(z) = $\displaystyle \frac{1}{z^2 - z - 6}$, $\displaystyle z_o = 1$

My attempt

i) I expanded the function as follows:

$\displaystyle \frac{1}{(z - 3)(z + 2)}$

Clearly there's singularities at z = 3, -2

I draw a quick graph and noticed that the radius of convergence is |z - 1| < 2

This part I got correct. However, I get stuck for parts ii) and iii)

ii) I used partial fractions to get:

$\displaystyle \frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2}$

I then manipulated each fraction so I could write them in terms of known Macluarin series $\displaystyle \frac{1}{z \pm 1}$:

$\displaystyle \frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n)$ and

$\displaystyle \frac{-1}{15}(\sum_{n = 0}^{\infty}{(-1)^n(\frac{z - 1}{3})^n)$

Thus I got this for the series:

$\displaystyle \frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n) \frac{-1}{15}(\sum_{n = 0}^{\infinity}{(-1)^n(\frac{z - 1}{3})^n)$

Now, upon substituting n = 0, 1, 2 to get the first three terms, I get the following:

$\displaystyle -\frac{1}{5}$, $\displaystyle \frac{13(z - 1)}{180}$, $\displaystyle \frac{19(z - 1)^2}{1080}$

These are wrong, and I'm not sure why. :/

Any assistance would be appreciated.