Find the Laurent expansion of: $$\frac{i}{z+i}$$
What should I do?
Do I need to times it by the complex conjugate? In that case I get:
$$\frac{i}{z+i}=\frac{iz+1}{z^2+1}$$ , I then tried to complete the square:
$$\frac{iz+1}{z^2+2z-2z+1}=\frac{iz+1}{(z+1)^2-2z}$$, which doesn't get me anywhere. On the other hand if I keep it imaginary, I don't know any power series expansions to go with them.
Say $\;\left|\frac zi\right|=|z|<1\;$ , then
$$\frac i{z+i}=\frac1{1+\frac zi}=\sum_{n=0}^\infty\frac{(-1)^nz^n}{i^n}$$
or if $\;\left|\frac iz\right|=\frac1{|z|}<1\iff |z|>1\;$ , then
$$\frac i{z+i}=\frac iz\frac1{1+\frac iz}=\frac iz\sum_{n=0}^\infty\frac{(-1)^ni^n}{z^n}$$
About the only complex pole, i.e. $\;z=-i\;$ , the series is pretty boring: it is $\;\frac i{z+i}\;$ .