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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

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(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) $|z|>3$

\begin{align*} & \frac{5}{z+2} - \frac{10}{z+3} \\ =& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{z} \frac{1}{1+\frac{3}{z}} \\ =& \frac{1}{z} \sum_{n = 0}^\infty (-1)^n \left( 5 \left( \frac{2}{z} \right)^n - 10 \left( \frac{3}{z} \right)^n \right) \\ =& \sum_{n = 1}^\infty (-1)^{n - 1} \left( 5 \left( \frac{2}{z} \right)^n - 10 \left( \frac{3}{z} \right)^n \right) \end{align*}

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  • $\begingroup$ The only issue I'm still having is for part (i): You use the sum version of 10/(z+3) even though it's only valid for mod(z) < 3, when the question wants the series for mod(z) < 2. Why is it OK to do this? (Similarly for part (iii) - you use the sum version of 5/(z+2) even though it's only valid for mod(z) > 2, and the question asks for the region mod(z) > 3) $\endgroup$ – emc3636 Dec 22 '15 at 16:41
  • $\begingroup$ (i) $|z| < 2 < 3$ (iii) $|z| > 3 > 2$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 22 '15 at 16:45

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