Laurent expansion correct?

Question

I'm trying to expand $\frac{z^2}{1-z}$ in order that I can find its residue. I expanded $\frac{1}{1-z}$ first, and then multiplied that expansion by $z^2$. I got: $-z-1-\frac{1}{z^2}-...$ Can someone tell me if this is correct, because it's just not looking right to me...

Answer 1

(i) expansion around $z=1$ is $$\frac {1+(z-1)}{-(z-1)}=-\frac{1}{z-1}-1$$(ii) expansion around $z=0$ is $$z(1+z+z^2+z^3+...)=z+z^2+z^3+z^4+...$$

Answer 2

I am going to provide some details to P.Lawrence's answer.

When you say the Laurent expansion of $\frac{z^2}{1-z}$, you should first say at which point you are expanding the function. If you expand the function at $z=0$, then you should get a Laurent (or power) series like this $\sum_{n \in \mathbb{Z}} a_n z^n$, though it may contain finitely many or zero negative terms. When you expand the function at $z=1$, you should anticipate something like $\sum a_n (z-1)^n$. These two series are not the same. If you expand at $z=0$, you will get $z^2+z^3+\dots$, which P.Lawrence wrote. This series only converges when $|z|<1$. If you substitute $z=1$ into the series, you will get $\infty$.

Therefore, the point you expand at determines the radius of convergence. The function $\frac{z^2}{1-z}$ is holomorphic except at the point $z=1$. In complex analysis, being holomorphic equals to being analytic. So this means at any point $z \neq 1$, you can have an open ball $B$ centered at $z$, such that you can have a power series expansion at $z$ that converges in that open ball $B$. In general, this open ball can be quite large as long as it doesn't contain any singularities. In your case, if you expand the function at $z=0$, the largest ball it can take is the open unit ball because it will touch the singularity $z=1$ if you enlarge the ball one little step more.

So the series $z^2+z^3+\dots$ succeeds to represent $\frac{z^2}{1-z}$ until the boundary of the open ball touches the singularity $z=1$. But this is not satisfying for us. There is only one singularity and the function $\frac{z^2}{1-z}$ is still holomorphic after $z=1$ onwards. Can we expand $\frac{z^2}{1-z}$ at $z=0$ and got a series working even for $|z|>1$? The answer is yes.

Let $\frac{1}{1-z} = \frac{1}{z} \frac{-1}{1-1/z}$. If you think of $\frac{1}{z}$ as a single variable, then $\frac{1}{1-1/z}=1+1/z+1/z^2+\dots$. So finally you got $\frac{1}{1-z}=-\frac{1}{z}-\frac{1}{z^2}-\dots$ when $|z|>1$. Then $\frac{z^2}{1-z}=-z-1-\frac{1}{z}-\dots$ when $|z|>1$.