So I'm trying to calculate the Laurent series of $f(z)=\frac{1}{2z^3}-\frac{2}{z^3+i}$.

Here are the steps I beieve I should take;


Well first I noted that it has two singularities $z_0=0$ and $z_0=\sqrt[3]{-i}$.

And I believe that because there are epsilon neighbourhoods around them both that do not intersect that this means they are isolated singularities and so we must calculate two Laurent series 1 for each singularity.

Step 2)

consider first the singularity $z_0=\sqrt[3]{-i}$.

Try o calculate a series for the term which contains this singularity, namely$ -\frac{2}{z^3+i}$. this can be done by rearranging until we have a form which can be expressed a geometric series.

We then add $1/2z^3$ (evaluated at $z=\sqrt[3]{-i}$ )to the begining of the geometric series obtained and say that this is the laurent series for z=i

step 3)

we know have to classify the singularity the singularity at $z_0=0$. We do this by calculating the taylor series of $2z^3$ and then noting that $\frac{2z^3}{2z^3}=1$ and use this fact to find the terms of $\frac{1}{2z^3}$. we then evaluate $\frac{-2}{z^3+i}$ at $z=0$ and add this to our series which we say is our Laurent series for z=0

So is this correct method for computing Laurent series of this form ?

Furthermore is there anything else I should have noted here and if it is about right , is there any way I could have improved on my method ?

  • $\begingroup$ Doesn't $-i$ have 3 cube roots? $\endgroup$ May 17, 2018 at 21:24
  • $\begingroup$ @AlexPavellas oh shoot you're right , so then I should really have expanded that term out to include the three roots and done a partial fractions shouldn't I ? Also I thinkI made another mistake , namely that I said that I said we should add the term we didn't do a laurent series for evaluated at the singularity of the one we did compute the singularity , which I believe is incorrect after working on doing examples all day ? $\endgroup$
    – excalibirr
    May 17, 2018 at 21:31

1 Answer 1


Notice that $-\frac{2}{z^3+i}$ has a Taylor series about $0$, further notice that $\frac{1}{z^3}$ is its own Laurent Series. What does this tell you about $\frac{1}{z^3}-\frac{2}{z^3+i}$?


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