I came across to calculate an integral for which I had to find the Laurent series of $f(z)=\dfrac{\sqrt{z}}{z+i}$.
I see that at $z=-i$ the function has a simple pole and the Laurent series I got is as follows:
for $0<|z|<1$ we have $\dfrac{\sqrt{z}}{i}\,\left(\dfrac{1}{1+\dfrac{z}{i}} \right)=\dfrac{\sqrt{z}}{i}\,\left(1-\dfrac{z}{i}+\dfrac{z^2}{i^2}-\cdots \right)$
and for $|z|>1$ we have $\dfrac{\sqrt{z}}{z}\,\left(\dfrac{1}{1+\dfrac{i}{z}} \right)=\dfrac{\sqrt{z}}{z}\,\left(1-\dfrac{i}{z}+\dfrac{i^2}{z^2}-\cdots \right)$.
Thus $f(z)=\dfrac{\sqrt{z}}{i}\,\left(1-\dfrac{z}{i}+\dfrac{z^2}{i^2}-\cdots \right)+\dfrac{\sqrt{z}}{z}\,\left(1-\dfrac{i}{z}+\dfrac{i^2}{z^2}-\cdots \right)$.
I am not particularly sure about the answer. Any suggestions or hints would be appreciated.