I have observed that Series in Mathematica assumes that the given function is smooth in the point around which one wants to perform series expansion.

For instance:

Series[1/z[x]^2, {x, 0, 1}]

results in familiar Taylor expansion.

SeriesData[x, 0, {z[0]^(-2), (-2) z[0]^(-3) Derivative[1][z][0]}, 0, 2, 1]

whereas if we inform Mathematica about existence of pole of given function:

Assuming[{z[0] == 0}, Series[1/z[x]^2, {x, 0, 1}]]

Mathematica gives us something which looks like Laurent series:

SeriesData[x, 0, {Derivative[1][z][0]^(-2), -Derivative[1][z][0]^(-3) 
Derivative[2][z][0], Rational[1, 12] Derivative[1][z][0]^(-4) (9 Derivative[2][z][0]^2 - 
4 Derivative[1][z][0] Derivative[3][z][0]), Rational[1, 12]
Derivative[1][z][0]^(-5) ((-6) Derivative[2][z][0]^3 + 6 Derivative[1][z][0]
  Derivative[2][z][0] Derivative[3][z][0] - Derivative[1][z][0]^2 Derivative[4][z][0])}, -2, 2, 1]

Now, I wanted to convince myself that this is indeed the proper Laurent series, so I tried explicit integration:

1/(2 π I) Integrate[(1/z[Exp[I α]]^2)*(α)^2, {α, 0, 2 π}]

taking the contour to be unit circle. Unfortunately, Mathematica is not able to reproduce the coefficient form mentioned series. So my questions are:

  1. How to compute such complex integral?
  2. How does Series deal with such a problem?
  • $\begingroup$ According to the documentation of Series (under "Scope"), it does generate a Laurent series about poles. $\endgroup$
    – bbgodfrey
    Mar 11, 2015 at 0:22
  • 1
    $\begingroup$ You could explicitly define z[0]=0 and then take the Series. $\endgroup$ Mar 11, 2015 at 1:43

1 Answer 1


I don't know any way to tell Integrate that 1/z[x]^2 has a pole at zero. However, we can integrate your series expression:

series = Assuming[{z[0] == 0}, Series[1/z[x]^2, {x, 0, 1}]];
1/(2 Pi I) Integrate[Normal[series] * I x /. {x -> Exp[I t]}, {t, 0, 2 Pi}]

This gives us


This makes sense, since Cauchy's integral formula is:

$$ \oint_\gamma \frac{f(z)}{(z-a)^n} dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a) $$

For the pole of order 1, $f(z) = -\frac{z''(0)}{z'(0)^3}$ is a constant, so the residue is equal to that expression (times $2\pi i$).

For the pole of order 2, $f(z) = \frac{1}{z'(0)^2}$, again a constant. However, since the value of this residue depends on $f^{(1)}(z)=0$, the residue is simply zero, and only the first residue contributes to the result.


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