CVS3

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Find the Laurent series for {\frac  {1}{z^{2}(z-1)}}\, about all its singular points.

The singularities are z=0,1\,.

Centered at the singular point z_{0}=0\, there are two possible Laurent series. One valid for 0<|z|<1\, and one for |z|>1\,.

In the domain 0<|z|<1\,, {\frac  {1}{z^{2}(z-1)}}=-{\frac  {1}{z^{2}}}\sum _{{n=0}}^{\infty }z^{n}\,

=-\sum _{{n=0}}^{\infty }z^{{n-2}}=-\sum _{{n=-2}}^{\infty }z^{n}\,

For the domain |z|>1\,,{\frac  {1}{z^{2}(z-1)}}={\frac  {1/z}{z^{2}(1-1/z)}}\,

={\frac  {1}{z^{3}}}\sum _{{n=0}}^{\infty }\left({\frac  {1}{z}}\right)^{n}=\sum _{{n=0}}^{\infty }\left({\frac  {1}{z}}\right)^{{n+3}}=\sum _{{n=0}}^{\infty }\left({\frac  {1}{z}}\right)^{{n+3}}\,

Let n\to -n-3\,.

=\sum _{{n=-\infty }}^{{-3}}z^{{n}},|z|>1\,

For the singular point z=1\, make a change of variables u=z-1\,.

{\frac  {1}{z^{2}(z-1)}}={\frac  {1}{(u+1)^{2}u}}\,

Now when 0<|u|<1\,

{\frac  {1}{(u+1)^{2}u}}={\frac  {1}{u}}\sum _{{n=0}}^{\infty }(-1)^{n}(n+1)u^{n}{\Bigg |}_{{u=z-1}}\,

={\frac  {1}{z-1}}\sum _{{n=0}}^{\infty }(-1)^{n}(n+1)(z-1)^{{n-1}}\,

=\sum _{{n=0}}^{\infty }(-1)^{n}(n+1)(z-1)^{{n-2}}\,

With a change of index n\to n+1\,

\sum _{{n=-1}}^{\infty }(-1)^{{n+1}}(n+2)(z-1)^{n},0<|z-1|<z\,

When |u|>1\,

{\frac  {1}{(u+1)^{2}u}}={\frac  {1}{u^{3}(1+1/u)^{2}}}={\frac  {1}{u^{3}}}\sum _{{n=0}}^{\infty }(-1)^{n}(n+1)u^{{-n}}{\Bigg |}_{{u=z-1}}\,

={\frac  {1}{(z-1)^{3}}}\sum _{{n=0}}^{\infty }(-1)^{n}(n+1)(z-1)^{{-n}}\,

=\sum _{{n=0}}^{\infty }(-1)^{n}(n+1)(z-1)^{{-n-3}}\,

=\sum _{{n=-\infty }}^{{-3}}(-1)^{n}(n+2)(z-1)^{n},|z-1|>1\,


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