# Laurent series in various regions

I have the question: "Find the Laurent series which represents the function

$$f(z) = (z^2 - 1)/(z + 2)(z + 3)\$$

in the regions

(i) $\mid z\mid < 2\$

(ii) $2 < \mid z\mid < 3\$

(iii) $\mid z\mid > 3\$"

I can rewrite this function into partial fractions as:

$$f(z) = 5/(z+2) - 10/(z+3)$$

I understand that for part (ii) this is the region within the annulus.

I also know that I can write:

$$5/(z+2) = 5/2 \sum (-1)^n (z/2)^n$$ for $\mid z\mid < 2\$

$$10/(z+3) = 10/3 \sum (-1)^n (z/3)^n$$ for $\mid z\mid < 3\$

I also know how to find the respective sums for $\mid z\mid > 2\$ and $\mid z\mid > 3\$ But I'm not at all sure how to combine these to get the answers for the whole function $f$, for parts (i), (ii) and (iii).

Any help would be much appreciated

(i) $|z| < 2$
$$\frac5{z+2}-\frac{10}{z+3} = \sum_{n=0}^\infty (-1)^n \left(\frac52\left(\frac{z}2\right)^n-\frac{10}3\left(\frac{z}3\right)^n\right)$$
(ii) $2<|z|<3$ \begin{align*} & \frac{5}{z+2} - \frac{10}{z+3} \\ =& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{3} \frac{1}{1+\frac{z}{3}} \\ =& -\frac{10}{3} \sum_{n = 0}^\infty (-1)^n \left( \frac{z}{3} \right)^n + \frac{5}{z} \sum_{n = 0}^\infty (-1)^n \left( \frac{2}{z} \right)^n \\ =& \sum_{n = 0}^\infty (-1)^{n + 1} \frac{10}{3} \left( \frac{z}{3} \right)^n + \sum_{n = 1}^\infty (-1)^{n - 1} 5 \left( \frac{2}{z} \right)^n \end{align*}
(iii) $|z|>3$
• (i) $|z| < 2 < 3$ (iii) $|z| > 3 > 2$ – GNUSupporter 8964民主女神 地下教會 Dec 22 '15 at 16:45