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I want to obtain the Laurent of the function f(z) = 1/(z(z-1)) valid first in 0 < Abs[z] < 1 and second in Abs[z] > 1. I tried

Series[f[z], {z, 0, 4}, Assumptions -> (0 < Abs[z] < 1)]

obtaining the correct result, but when I change the Assumptions

Series[f[z], {z, 0, 4}, Assumptions -> Abs[z] > 1]

I obtain the same result, which is wrong. What am I doing wrong?

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    $\begingroup$ Try Series[1/(z*(z - 1)), {z, Infinity, 4}]. $\endgroup$
    – user64494
    Commented Aug 11, 2021 at 15:51
  • $\begingroup$ It works. I ask however what's the use of Assumptions in series of complex functions... $\endgroup$
    – CharlesG
    Commented Aug 11, 2021 at 17:04

1 Answer 1

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Up to the documentation, the Assumptions option of the Series command works for parameters only. The required expansion can be produced by

Series[1/(z*(z - 1)), {z, Infinity, 4}]
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