Laurent proper circular rings How can I find all proper circular rings with centre $ z=-i $
to compute Laurent Series of $ f(z)=\frac{1}{z*(z-1)} $ ?
I thought $$u=z+i$$ $$f(z)=\frac{1}{z*(z-1)}=\frac{1}{z*(z+i-i-1)}$$ 
$$f(u)=\frac{1}{(u-i)*(u-i-1)} $$
So the proper circular rings are
$$ |u| > 1+i $$
$$ |u| < i $$
$$ i<|u| <1+i $$
??
 A: The function
\begin{align*}
f(z)&=\frac{1}{z(z-1)}\\
&=\frac{1}{z-1}-\frac{1}{z}\\
\end{align*}
is to expand around the center $z=-i$.
Informally: In order to determine the regions of convergence in case of isolated singularities we start from the point we want to expand the function (here $z=-1$). Then we look at the greatest circular region which does not contain a singularity and so has necessarily a point of singularity at its boundary. This determines the first region. Then we consider an annulus starting from the first point of singularity circularly extended to the next point of singularity. We continue this way till all singularities are at the boundary of one of these regions. The final region is the complement of (the union of the closure) of all the other regions.

Since there are simple poles at $z=0$ and $z=1$ we have to distinguish three regions of convergence
  \begin{align*}
D_1:&\quad 0\leq  |z+i|<1\\
D_2:&\quad 1<|z+i|<\sqrt{2}\\
D_3:&\quad  |z+i|>\sqrt{2}
 \end{align*} 
  
  
*
  
*The first region $D_1$ is a disc with center $z=-i$, radius $1$ and the pole at $z=0$ at the boundary of the disc. It admits for both fractions a representation as power series. 
  
*The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It has the poles $z=0$ and $z=1$ at the boundary of the annulus. It admits for the fraction with pole  at $z=0$ a representation as principal part of a Laurent series and for the fraction with pole at $z=1$ a representation as power series.
  
*The region $D_3$ contains all points outside the disc with center $z=-i$ and radius $\sqrt{2}$. It admits for both fractions a representation as principal part of a Laurent series.

Hint: Note that in order to determine the radius of convergence we use the total ordering of the real numbers. So, $|u|<i$ is not feasible, since $i\not\in\mathbb{R}$. We have to consider $|u|<|i|=1$ instead.
