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I wanna check my laurent series exercises on Mathematica, but can't seem to find a command or program to achieve the result of such type of interval.

$f(z)=\frac{1}{(z-3) z},\\1<|z-3|<3$

The command Series[] won't let me use $z-3$ as the expansion variable.

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If you want to expand your series about the point z = 3 you can do this:

Series[1/(z - 3)/z, {z, 3, n}]

Just change n to be the number of terms you want in the expansion.

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    $\begingroup$ You can use SeriesCoefficient too and not need to replace n with an explicit integer. $\endgroup$ – Chip Hurst Oct 18 '16 at 17:45
  • $\begingroup$ But wouldn't it be the same as calculate the series for 0 < |z| <3? $\endgroup$ – Vinholi Oct 18 '16 at 18:42

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