# Series development of laurent in a defined domain

I am trying to correct some bills for laurent series with mathematica, but the output I am getting at the moment is not the best. For example, I have this function $$\frac{1}{z^2 + 9}$$ to develop at the point $$z_0 = 3i$$ in region $$0 < |z - 3i| < 6$$.

What I have tried so far is

Series [1 / (z ^ 2 + 9), {z, 3 I, 10}]


but this only returns me 10 terms of the series.

Then I tried

sumRule =
Inactive[Series][f_, {x_, x0_, n_}] :>
Inactive[Sum][
Assuming[{Element[k, Integers], k >= 0},
SeriesCoefficient[f, {z, 3 I, k}] (x - x0)^k //
FullSimplify], {k, 0, n}];

n = Infinity;

f[z_] = 1/(z^2 + 9);

Inactive[Series][f[z], {z, 3 I, n}] /. sumRule


The latter returns $$\underset{k=0}{\overset{\infty }{\sum }}6^{-k-2} (3+i z)^k$$ which comes close to the expected result: $$\frac{1}{6i} \frac{1}{z-3i} + \sum_{n=0}^{+\infty} \frac{-1}{(6i)^{n+2}} (z-3i)^n$$ but it is not this complete and I don't know why. I also haven't specified anywhere in which region to develop. So how can I get the correct result and maybe even specify the region?

• "but this only returns me 10 terms of the series" What exactly do you mean by this? I am asking because you did ask for 10 terms? Jan 19, 2022 at 3:07
• Because it would serve me up to infinity, but rightly does not print infinite terms.
– Teo7
Jan 19, 2022 at 9:56

\$Version

(* "13.0.0 for Mac OS X x86 (64-bit) (December 3, 2021)" *)

Clear["Global*"]

c[k_] = SeriesCoefficient[1/(z^2 + 9), {z, 3 I, k}]


sum = Inactive[Sum][c[k]*(z - 3*I)^k, {k, -1, Infinity}] //
Simplify[#, {k >= -1, Element[k, Integers]}] &


Verifying,

sum // Activate

(* 1/(9 + z^2) *)


EDIT: Your "expected result" does not equal the original function

sum2 = HoldForm[1/(6 I) 1/(z - 3 I) +
Inactive[Sum][-1/((6 I)^(n + 2)) (z - 3 I)^n, {n, 0, Infinity}]]


sum2 // ReleaseHold // Activate // Simplify

(* (6 + I z)/(81 + 36 I z - 3 z^2) *)
`
• Thank you very much for the reply and for the correction. However, this procedure does not seem universal, for example SeriesCoefficient does not seem to work with the function (z ^ 2 (z - 1)) / (Sin [1 / (z - 1)]) and I don't know why.
– Teo7
Jan 19, 2022 at 10:42