Laurent series with imaginary numbers I've been trying to understand how to solve this Laurent series with partial fractions and substitutions, but I can't seem to get it right;
$$0<|z-2i|<R$$
$$f(z) = \frac{1}{z^2+4} = \frac{1}{(z-2i)(z+2i)}$$
I'm having a hard time understanding how to work with the zero, and also how to calculate the biggest value that $R$ can get.
 A: Le $w:=z-2i$, then for $0<|w|<4$, we have that
$$f(z)=\frac{1}{w}\cdot \frac{1}{w+4i}=\frac{1}{4iw}\cdot \frac{1}{1-\frac{iw}{4}}=-\frac{i}{4w}\cdot \sum_{k=0}^{\infty}\left(\frac{iw}{4}\right)^k\\
=-\sum_{k=0}^{\infty}\left(\frac{i}{4}\right)^{k+1} w^{k-1}
=-\sum_{k=-1}^{\infty}\left(\frac{i}{4}\right)^{k+2} (z-2i)^{k}
$$
A: We can use partial fraction expansion to write $f(z)$ as 
$$f(z)=\frac{1}{i4}\left(\frac{1}{z-i2}-\frac{1}{z+i2}\right)$$
Note that the poles are separated by a distance $|i2-(-i2)|=4$.  Therefore, let's write the Laurent series in the circle $|z-i2|<4$.  The first term in that series is $\frac{1/i4}{z-i2}$.  Let's expand the second term around $i2$.
$$\begin{align}
\frac{1}{z+i2}&=\frac{1}{(z-i2)+i4}\\\\
&=\frac{1}{i4}\frac{1}{1+\frac{z-i2}{i4}}\\\\
&=\frac{1}{i4}\sum_{n=0}^\infty (-1)^n \left(\frac{z-i2}{i4}\right)^n\\\\
&=-\sum_{n=0}^\infty \left(\frac{i}{4}\right)^{n+1} (z-i2)^n
\end{align}$$
Finally, we can write
$$f(z)=- \sum_{n=-1}^\infty \left(\frac{i}{4}\right)^{n+2} (z-i2)^n$$
