I have this function $f(z)=\frac{1-iz}{1+iz}$ and I would like to find the Laurent expansion around the singularities of it.
So first I know there is a singuality at $1+iz=0$ so $z=i$. Next rationalising the denominator I get:
$$f(z)=\frac{1-z^2-2iz}{z^2+1}$$.
This has singularities at $z=±i$. Then I am not sure how to further. Next say we look at $z=i$, so:
$$f(z)=\frac{1-z^2-2iz}{(z-i)(z+i)}$$
Then about our singularity:
$$f(z)=\left(\frac{1}{z-i}\right)\frac{1-z^2-2iz}{(z+i)}$$
$$f(z)=\left(\frac{1}{z-i}\right)\frac{(1-iz)(1+iz)}{(z+i)}$$
However I am not sure how to continue this? Could anyone help me out?
The only singularity is located at $z=i$ and you have$$\frac{1-iz}{1+iz}=\frac{z+i}{-z+i}=-1-2i(z-i)^{-1}.$$