I'm kind of stumped on a question here.
I've been asked to determine and classify the singularities of;
$$f(z) = \frac{z^3}{(1+z)^3}$$
To me, it's pretty obvious that a singularity will occur when $z = -1$, however, I'm now having trouble being able to classify this, because I'm not sure how to go about determining my expansion.
I was given the hint in class to essentially "add zero", to make the function look more like a Laurent series, but I'm really not confident with what I'm meant to be looking for here.
Any help would be fantastic.
Best Answer
Note that for all complex numbers $z$ with $z\neq -1$, $\dfrac 1{(z+1)^3}=\dfrac 1{(z-(-1))^3}$ so if you find the laurent series (which will just be the taylor series) of $z\mapsto z^3$ around $-1$, multiplying the series by $\dfrac 1{(z-(-1))^3}$ will yield the laurent series of $f$ at $-1$.
This is what the hint is getting at, more specifically, add $0$ in the following way: $$z^3=((z+1)-1)^3=(z+1)^3-3(z+1)^2+3(z+1)-1.$$
One identifies the RHS as the taylor series of $z\mapsto z^3$ around $-1$ and so the laurent series of $f$ around $-1$ is $$-\dfrac 1{(z+1)^3}+\dfrac 3{(z+1)^2}-\dfrac 3{z+1}+1.$$
However this is all unnecessary to answer the question. One can immediately see that $$\lim \limits_{z\to -1}\left((z+1)^3\dfrac{z^3}{(1+z)^3}\right)\in \mathbb C\setminus \{0\}$$ and $$\left|\lim \limits_{z\to -1}\left((z+1)^2\dfrac{z^3}{(1+z)^3}\right)\right|=\infty.$$
Thus $-1$ is a pole of order $3$ of $f$. It is $f$'s only singularity.