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Expand in a laurent Series : 1- $f_{1} (z) = \frac{z^{2} - 2z +5 }{(z^{2}+1) (z-2)}$

in the ring : $1 < |z| < 2 $

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

In : $i. 2 < |z| < 3 \\ ii. 0 < |z+2| < 5$

I managed to solve the second one but not sure if it is correct For i. $2 < |z| < 3$ :

$ \frac{-1}{5} * \frac{1}{z(1+ \frac{2}{z}) } + \frac{1}{5} * \frac{1}{-3(1- \frac{z}{3}) } = \frac{-1}{5} \sum_{n=0}^ \infty (-1)^{n} (\frac{2}{z})^{n} - \frac{1}{15}\sum_{n=0}^ \infty (\frac{z}{3})^{n}$

For ii. $0 < |z+2| < 5$ : $ \frac{-1}{5} * \frac{1}{z+2} + \frac{1}{5} * \frac{1}{(z+2 -5) } = \frac{-1}{5} * \frac{1}{z+2} + \frac{1}{25} * \frac{1}{-5 (1- \frac{Z+2}{5} ) } \\ = \frac{-1}{5} * \frac{1}{z+2}- \frac{1}{25} \sum_{n=0}^ \infty (\frac{z+2}{5})^{n} $

Radwa Kamal
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  • ...Please...? Did you try anything? – DonAntonio May 25 '13 at 09:05
  • Laurent, not Laurant. – Did May 25 '13 at 09:28
  • yes I tried but when I subtituted some values doesn't get proper results :/ I made Partial Fraction for the second one and for i : $\frac{- \frac{1}{5} }{z+2} + \frac{ \frac{1}{5} }{z-3}$ = $\frac{- \frac{1}{5} }{z(1+ 2/z) }$ + $\frac{- \frac{1}{5} }{-3(1- z/3)}$ ,, I Expanded both series and I guess this expansion is right but ii can't figure it out

    While f1 Can't find a way to figure it out :/

     – Radwa Kamal May 25 '13 at 09:48

1 Answers1

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In case 1, note that $1\lt|z|\lt2$ means that $1/|z|\lt1$ and $|z|/2\lt1$, hence the well known expansion $1/(1-u)=\sum\limits_{n\geqslant0}u^n$, valid for every $|u|\lt1$, applied twice, yields $$ \frac1{z^2+1}=\frac1{z^2(1+1/z^2)}=\sum\limits_{n\geqslant0}\frac{(-1)^n}{z^{2n+2}},$$ and $$ \frac1{z-2}=-\frac12\frac1{1-z/2}=-\sum\limits_{n\geqslant0}\frac{z^n}{2^{n+1}}. $$ The expansion of $f_1(z)$ as a Laurent series follows from the decomposition $$ f_1(z)=\frac1{z-2}-\frac2{z^2+1}.$$

Did
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