Find the Laurent series of $f(z)=\frac{z}{(z+1)^2}$ around $z_o=-1$, and $g(z)=z\exp(\frac1{z+i})$ around $z_o=-i$.
For $f$, what they're asking is to find the series in $0<|z+1|$. On the other hand, we have that: $$f(z)=\frac{z}{(z+1)^2}=\frac z{z+1}\frac 1{z+1}=\left(1-\frac1{z+1}\right)\frac1{z+1} \\=\frac1{z+1}-\frac1{(z+1)^2}=(z-z^2+z^3-z^4+\dots)-(1-2z+3z^2-4z^3+5z^4-\dots) \\=-1+3z-4z^2+5z^3-6z^4+\dots$$
But this series is around $0$, not $-1$, how do I change it? I was thinking in factorize but didn't worked. Then I thought maybe consider $0<|z+1|\le|z|+1\Rightarrow-1<|z|$, but I don't see where would the $-1$ be.
For $g$, as before we have $0<|z+i|$, and: $$g(z)=z\exp\left(\frac1{z+i}\right)=z\left[1+\frac1{z+i}+\frac1{2!}\frac1{(z+i)^2}+ \frac1{3!}\frac1{(z+i)^3}+\dots\right] \\=z+\frac z{z+i}+\frac1{2!}\frac z{(z+i)^2}+ \frac1{3!}\frac z{(z+i)^3}+\dots \\=z+\left(1-\frac i{z+i}\right) +\frac1{2!} \left(\frac 1{z+i}-\frac i{(z+i)^2}\right)+ \dots+\frac1{k!}\left(\frac 1{(z+i)^{k-1}}-\frac i{(z+i)^k}\right)+\dots \\=z+1+\frac 1{z+i}\left(-i+\frac1{2!}\right)+\frac 1{(z+i)^2}\left(\frac1{3!}+\frac{-i}{2!}\right)+\dots+\frac 1{(z+i)^k}\left(\frac1{(k+1)!}+\frac{-i}{k!}\right)+\dots$$
However, the first two terms are bothering me a lot, I can't seem to make them fit with the general term $\frac 1{(z+i)^k}\left(\frac1{(k+1)!}+\frac{-i}{k!}\right)$ even with $k<0$. Am I doing something wrong? is the way I'm doing this wrong?
You were actually done much earlier with $f.$ Its Laurent series about $z=-1$ is simply $$\frac1{z+1}-\frac1{(z+1)^2}.$$
As for $g,$ just rewrite $z+1=z+i+1-i.$ So, the coefficient of $(z+i)^1$ is $1,$ the coefficient of $(z+i)^{-k}$ for $k\ge 0$ is as you've already determined.