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How do I expand a function $f(z)$ in a particular region? For example, how would I expand $f(z)=(z^2-3z+2)^{-1}$ in the region $0<|z-1|<1.$? I believe this can be done by the binomial theorem. But how to arrange things to make the binomial expansion valid?

Please give me an idea for doing this.

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  • $\begingroup$ Is this a Mathematica question or a Math question? $\endgroup$
    – bobthechemist
    Commented Mar 21, 2014 at 13:31

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In Mathematica:

Series[1/(z^2 - 3 z + 2), {z, 1, 10}, Assumptions -> {0 < Abs[z - 1 ]< 1}]

(or whatever the top-order term is that you want).

Mathematical method:

Step (i): expand the function into partial fractions -- in Mathematica:

Apart[1/(z^2 - 3 z + 2)]

One of the fractions you get is the the Laurent series term with negative power of $z - 1$.

Step (ii): Expand the other term using a binomial series.

Step (iii): Add the results of Steps (i) and (ii).

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  • $\begingroup$ okay thanks.What to do if the region is |z|<1 ? $\endgroup$
    – user3185198
    Commented Mar 21, 2014 at 17:27
  • $\begingroup$ For $|z| < 1$, you'll need to expand around $z = 0$, not $z = 1$. And from the code I provided, I think you can see how to express that in Mathematica. $\endgroup$
    – murray
    Commented Mar 21, 2014 at 19:38
  • $\begingroup$ okay thanks.Can u explain the general case? $\endgroup$
    – user3185198
    Commented Mar 22, 2014 at 3:36
  • $\begingroup$ Not sure what you mean by "the general case": how expand around an arbitrary point for this one particular function, or how to treat any rational function's expansion, or something even more general? In any case, you need to know whether the region includes a singularity, especially a pole, which would make it like the situation of expanding around $z = 1$; if not, then you want a Taylor series around some point in the region. Any standard book about complex analysis will help with these things. $\endgroup$
    – murray
    Commented Mar 22, 2014 at 15:07

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