Laurent Series of $f(z) = \frac{(z+1)^2}{z(z^3+1)}$ about $z = 0$?

Question

This is supposed to be a non-calculator question, so I managed to get this far;

• $ z^3 + 1 = (z + 1)(z^2 - 1 +1)$ , by polynomial division.

• Therefore, $ f(z) = \frac{z+1}{z(z^2-z+1)} = (1 + \frac{1}{z})(z^2-z+1)^{-1}$

Now at this stage, my thinking is that I need to expand the second term into partial fractions. But it gets terribly messy. My other mode of thought is to just do this...

• $ \frac{1}{z^3+1} = \sum(-1)^{n}(z)^{3n}$

And to continue simply on from there. But I am unsure which way is the right way?

Answer 1

You have$$\frac1{1+z^3}=1-z^3+z^6-z^9+\cdots$$if $|z|<1$. So,$$\frac z{1+z^3}=z-z^4+z^7-z^{10}+\cdots$$(again, if $|z|<1$) and therefore$$\frac{z+1}{z^3+1}=1+z-z^3-z^4+z^6+z^7-\cdots$$It follows from this that\begin{align}\frac{(z+1)^2}{z^3+1}&=\frac{z+1}{z^3+1}(z+1)\\&=(1+z-z^3-z^4+z^6+z^7-\cdots)(1+z)\\&=(1+z-z^3-z^4+z^6+z^7-\cdots)+(z+z^2-z^4-z^5+z^7+z^8-\cdots)\\&=1+2 z+z^2-z^3-2 z^4-z^5+z^6+2z^7+\\&\qquad+z^8-z^9-2 z^{10}-z^{11}+z^{12}+2 z^{13}+\cdots\end{align}So, the Laurent series that you are after is$$\frac1z+2+z-z^2-2z^3-z^4+z^5+2z^6+z^7-z^8-2z^9-z^{10}+z^{11}+2z^{12}+\cdots$$