Classification of singularities around $i$ I need to classify singularities of : $$ f(z)=\frac{1}{z(1+z^2)}$$
if i do that around $0$ it no problem to find the Laurent series. 
but , im asked to do it around $i$.
Im having trouble finding the way to express the function as $|z-i|$ terms.
 A: The function 
$$f(z)=\frac{1}{z(1+z^2)}$$
has singularities at : 
$$z(1+z^2)=0 \Rightarrow \begin{cases} z=0 \\ z = \pm i\end{cases}$$
As a standard way in finding Laurent Series, let's manipulate the given function to fit the singularity around $i$ : 
$$f(z)=\frac{1}{z(1+z^2)} = \frac{1}{z(z-i)(z+i)}= \frac{1}{z(z-i)(z-i+2i)}=\frac{1}{z(z-i)(z-i)(1+\frac{2i}{z-i})}$$
$$=$$
$$\frac{1}{z(z-i)^2(1+\frac{2i}{z-i})}$$
A common geometric series, is : 
$$\frac{1}{1+w} = \sum_{n=0}^\infty (-1)^nw^n, \space |w| <1 $$
which means that the given function can be written as : 
$$f(z) = \frac{1}{z(z-i)^2}\sum_{n=0}^\infty(-1)^n\frac{(2i)^n}{(z-i)^n}=\frac{1}{z}\sum_{n=0}^\infty(-1)^n\frac{(2i)^n}{(z-i)^{n+2}}$$
with the constraint of : 
$$\bigg| \frac{2i}{z-i}  \bigg| < 2 \Leftrightarrow |z-i| < 1$$
Can you now finalize the expression of $f(z)$ ?
A: Let $u = z-i$. Then:
$$\begin{array}{rcl}
f(z) &=& f(u+i) \\
&=& \displaystyle \frac 1 {u(u+i)(u+2i)} \\
&=& \displaystyle \frac i u \left[ \frac 1 {u+2i} - \frac 1 {u+i} \right] \\
&=& \displaystyle \frac i u \left[ \frac {-0.5i} {1-0.5iu} - \frac {-i} {1-iu} \right] \\
&=& \displaystyle \frac 1 u \left[ \frac {0.5} {1-0.5iu} - \frac {1} {1-iu} \right] \\
&=& \displaystyle \frac 1 u \left[ 0.5 \sum_{n=0}^\infty (0.5iu)^n - \sum_{n=0}^\infty (iu)^n \right] \\
&=& \displaystyle \frac 1 u \left[ \sum_{n=0}^\infty (0.5^{n+1}-1) i^n u^n \right] \\
\end{array}$$
A: According to the right-hand side of
\begin{align*}
\frac{1}{z(1+z^2)}=\frac{1}{z(z+i)(z-i)}
\end{align*}
we see there are three singularities, namely simple poles at $0,-i$ and $i$.

In order to obtain a Laurent-series expansion around $z=i$ we do at first a partial fraction expansion and
  obtain
  \begin{align*}
\frac{1}{z(z+i)(z-i)}=\frac{A}{z}+\frac{B}{z+i}+\frac{C}{z-i}
\end{align*}
  with $A,B$ and $C$ appropriate constants in $\mathbb{C}$.

We observe that $\frac{C}{z-i}$ is the principal part of the Laurent series expansion at $z=i$ while the other two terms are analytic at $z=i$ and can be expanded as power series.

Each of these terms can be expanded around $z=i$ according to
  \begin{align*}
\color{blue}{\frac{1}{z+a}}&=\frac{1}{(a+i)+(z-i)}=\frac{1}{a+i}\cdot\frac{1}{1+\frac{z-i}{a+i}}\\
&=\frac{1}{a+i}\sum_{n=0}^{\infty}\left(-\frac{z-i}{a+i}\right)^n\\
&\color{blue}{=\sum_{n=0}^{\infty}\frac{(-1)^n}{(a+i)^{n+1}}(z-i)^n}\\
\end{align*}

