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$\textbf{Exercise :}$ Find the Laurent series of :

$$ \dfrac{4z-z^2}{(z^2-4)(z+1)} $$

in the ring centered at $0$ and radius $1$ and $2$.

$\textbf{Solution :}$ Write for $1 < \vert z \vert < 2$ :

$$ \dfrac{4z-z^2}{(z^2-4)(z+1)} = \dfrac{1}{3(z-2)} + \dfrac{5}{3(z+1)} - \dfrac{3}{z+2} $$

We have :

$$\dfrac{1}{(z-2)} = -\dfrac{\dfrac{1}{2}}{1-\dfrac{z}{2}} = -\dfrac{1}{2} \sum_{n=0}^{\infty} (\dfrac{z}{2})^n$$

$$ \dfrac{1}{1+z} = \dfrac{\dfrac{1}{z}}{1-(-\dfrac{1}{z})} = \sum_{n=0}^{\infty} \dfrac{1}{z} (-\dfrac{1}{z})^n$$

$$ \dfrac{1}{2+z} = \dfrac{\dfrac{1}{2}}{1-(-\dfrac{z}{2})} = \sum_{n=0}^{\infty}\dfrac{1}{2} (-\dfrac{z}{2})^n$$

So : $$ \dfrac{4z-z^2}{(z^2-4)(z+1)} = \sum_{n=0}^{\infty} -\dfrac{1}{6}(\dfrac{z}{2})^n + \sum_{n=0}^{\infty} \dfrac{5}{3} \dfrac{(-1)^n}{z^{n+1}}- \sum_{n=0}^{\infty} \dfrac{3}{2}(-\dfrac{z}{2})^n$$

$$ \dfrac{4z-z^2}{(z^2-4)(z+1)} = \sum_{n=0}^{\infty} -\dfrac{1}{6.2^n}z^n + \sum_{n=0}^{\infty} \dfrac{5(-1)^n}{3} \dfrac{1}{z^{n+1}}- \sum_{n=0}^{\infty} \dfrac{3(-1)^n}{2^{n+1}}z^n$$

Is correct?, thanks for the help!

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  • $\begingroup$ Yes, logically it seems to be correct. $\endgroup$ – Dr Zafar Ahmed DSc Nov 26 '19 at 5:51
  • $\begingroup$ Is there any significance to why you thought of $\dfrac{1}{1+z} = \dfrac{\dfrac{1}{z}}{1-(-\dfrac{1}{z})}$ instead of just $\dfrac{1}{1+z} =\dfrac{1}{1-(-z)}$? I don't have much complex analysis experience, just wondering. $\endgroup$ – Biggs Nov 26 '19 at 6:01
  • $\begingroup$ @Biggs I presume because the OP wanted (correctly) an expansion in positive powers of $1/z$ rather than of $z$. $\endgroup$ – Angina Seng Nov 26 '19 at 6:19

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