Say we have the function $f(z)=\frac{1}{z^2+1}$ and we want to calculate it's Laurent series about $z=i$.
Then I know that we can decompose this function using partial fractions into
$f(z)=\frac{\frac{-1}{2i}}{z+}+\frac{\frac{1}{2i}}{z-i}$.
We then compute a geometric series for the one of the these which doesn't have a singularity for $z=i$.
$\frac{\frac{-1}{2i}}{z+i}=\frac{\frac{-1}{2i}}{(z-i)+2i}\frac{\frac{1}{2i}}{\frac{1}{2i}}=\frac{\frac{1}{4}}{1+\frac{(z-i)}{2i}}$
The geometric series is then given as $\frac{\frac{1}{4}}{1+\frac{(z-i)}{2i}}=\frac{1}{4}\sum_{n=0}^{\infty}(\frac{z-i}{2i})^n$.
Now in rearranging this sum , this is what i did ;
$\frac{1}{4}\sum_{n=0}^{\infty}\frac{(z-i)}{2i} = \sum_{n=0}^{\infty}(\frac{z-i}{2i})^n\frac{1}{4}=\sum_{n=0}^{\infty}(\frac{(z-i)^n}{(2i)^n}\frac{1}{2^2}=\sum_{n=0}^{\infty}\frac{(z-i)^n}{2^{n+2}i^n}=\sum_{n=0}^{\infty}(z-i)^n2^{-n-2}i^{-n}$
But in the example I saw this last bit was given as $=\sum_{n=0}^{\infty}(z-i)^n2^{-2n-2}i^{-n}$.
Did I make some kind of silly little mistake or was this a typo in the example ?