Laurent Expansion - Complex Analysis question Exercise :
Find the Laurent Expansion of the function :
$$f(z) = \frac{1}{(z-2i)(z^2 +4)}$$
around $z_0 = -2i$, in the biggest possible ring that includes the point $z=-2 + 2i$.
Attempt :
We have :
$$f(z) = \frac{1}{(z-2i)(z^2 +4)} = \frac{1}{(z-2i)(z + 2i)(z-2i)} = \frac{1}{(z-2i)^2(z + 2i)} $$
We have the rings :
$$D_1 : 0<|z+2i|<4$$
$$D_2 : |z+2i|>4$$
Because $|(-2+2i) + 2i| = |-2 + 4i| = \sqrt{20} > 4$, the biggest ring with center $z_0=-2i$ that contains the point $z=-2+2i$, is :
$$D_2 : |z+2i|>4$$
Now, such exercises can be handled by creating such a fraction, such as to use the usual geometric series : 
$$\frac{1}{1-w} = \sum_{n=0}^{\infty}w^n ,|w|<1$$
$$\frac{1}{1+w} = \sum_{n=0}^{\infty}(-1)^nw^n,|w|<1$$
It is : 
$$f(z) = \frac{1}{(z-2i)^2(z + 2i)} = \frac{1}{(z-2i)^2[(z - 2i)+4i]} = \frac{1}{(z-2i)^2(z - 2i)\big(1-\frac{4i}{z-2i}\big)} = \frac{1}{(z-2i)^3\big(1-\frac{4i}{z-2i}\big)}  $$
But in order to have $|w|<1 $ it must hold that : $|z-2i| > 4$, but our ring is $|z+2i|>4$. My question is : Is the exercise wrong ? Maybe it means around the center $z_0 = 2i$ ? Or is it something that I am doing wrong ? If nothing is wrong, how do I proceed here ?
Except if I use : 
$$\frac{1}{(z-2i)^2(z + 2i)} = \frac{1}{(z+2i)^3\big(1-\frac{4i}{z+2i}\big)^2} = \frac{1}{(z+2i)^3}\frac{1}{ \big(1-\frac{4i}{z+2i}\big)^2} = \frac{1}{(z+2i)^3} \Bigg( \sum_{n=0}^{\infty} \bigg(\frac{4i}{z+2i}\bigg)^n\Bigg)^2 $$
because if so, it is $|z+2i|>4$.
I would really appreciate some thorough help.
 A: You are on the right track. Just the last step should be revised a little.

The function $$f(z)=\frac{1}{(z-2i)^2}\cdot\frac{1}{(z + 2i)}$$ is to expand around the center $z_0=-2i$. Since there are poles at $z=2i$ and $z=-2i$ we have to distinguish two regions 
\begin{align*}
D_1:&\quad 0<|z+2i|<4\\
D_2:&\quad |z+2i|>4
 \end{align*} 
  
  
*
  
*The first region $D_1$ is a punctured disc with center $z_0=-2i$, radius $4$ and the pole $2 i$ at the boundary of the disc.
In the interior we have a representation of the fractions with a pole at $z=- 2i$ as principal part of a Laurent series at $z_0=-2i$, while the fraction with pole at $z=2i$ admits a representation as power series.
  
*The other region $D_2$ containing all points outside $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=-2i$.

We have to put the focus on $D_2$ since it contains the point $-2+2i$.

We  obtain 
  \begin{align*}
\color{blue}{f(z)}&\color{blue}{=\frac{1}{(z + 2i)(z-2i)^2}}\\
&=\frac{1}{(z + 2i)}\cdot\frac{1}{\left((z+2i)-4i\right)^2}\\
&=\frac{1}{(z + 2i)^3}\cdot\frac{1}{\left(1-\frac{4i}{z+2i}\right)^2}\tag{1}\\
&=\frac{1}{(z + 2i)^3}\sum_{n=0}^\infty \binom{-2}{n}\left(-\frac{4i}{z+2i}\right)^n\tag{2}\\
&=\sum_{n=0}^\infty (n+1)(4i)^n\frac{1}{(z+2i)^{n+3}}\tag{3}\\
&\color{blue}{=\sum_{n=3}^\infty (n-2)(4i)^{n-3}\frac{1}{(z+2i)^{n}}}\\
\end{align*}

Comment:


*

*In (1) we apply the binomial series expansion.

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (3) we shift the index to start with $n=3$.
