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;
Step.1)
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 ?
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 $\frac{1}{z^3}-\frac{2}{z^3+i}$?