I got the following question as part of some homework on Laurent Series, and I am about 90% sure that I am doing something wrong. Here is the question and my attempt at a solution:
Expand the function $$f(z) = \frac{z}{1+z^3}$$ (a) in a series of positive powers of z, and
(b) in a series of negative powers of z
In each case, specify the region in which the expansion is valid.
Like I said, I am 90% sure that I did something wrong, but here is my attempt at the problem:
(a) Consider the following function: $$f(z) = \frac{z}{1+z^3}$$ We know that each term in the positive expansion is $a_n = f^{(n)}(0)/n!$. Therefore, we find that the positive expansion is the following: $$f(z) = z-z^4+z^7-z^{10}+z^{13}-\ldots$$ This converges for all $|z| < 1$, since $f(z)$ is not analytic at $z = -1$.
(b) Consider the following function: $$f(z) = \frac{z}{1+z^3}$$ We know that each term in the Laurent series is the following, where $t < 1$ and $n \in \mathbb N^-$: $$a_n = \frac{1}{2\pi i}\int_{|z| = t}\frac{dz}{z^n(1+z^3)}$$ It is clear that $1/z^n(1+z^3)$ is analytic in $D(0, t)$. Therefore, each term in the sequence is the following from the Mean Value Theorem: $$a_n = 0$$ Therefore, the series of negative powers is the following: $$f(z) = 0$$
The first part is correct since
$$|z|<1\implies\frac1{1+z}=\sum_{n=0}^\infty(-1)^nz^n$$
and now you apply this to $\;z^3\;$ and then multiply by $\;z\;$ the whole thing and you get what you wrote.
For the second part I think it maybe better the following (as your answer is incorrect, I think):
$$|z|>1\implies\frac1{|z|}<1\implies\frac1{|z|^3}<1\;\;\implies\;\;\frac z{1+z^3}=\frac1{z^2}\cdot\frac1{1+\frac1{z^3}}=$$
$$=\frac1{z^2}\sum_{n=0}^\infty\frac{(-1)^n}{z^{3n}}=\sum_{n=0}^\infty\frac{(-1)^n}{z^{3n+2}}$$