I tried to expand three functions in Laurent series, but I don't have any given answer for them and I'm not very confident that what I'm doing is correct.
The first one is $f(z)=\frac{z^3e^{1/z}}{z+1}$ around $z_0 = 0$ and this is how I attempted to expand it:
$$f(z)=z^3e^{1/z}\frac{1}{1-(-z)}=z^3 \sum_{n=0}^{\infty} \frac{z^{-n}}{n!} \sum_{n=0}^{\infty} (-1)^{n} z^n =\sum_{n=0}^{\infty} \frac{(-1)^n z^3}{n!}$$
The second one is $f(z)=\frac{1}{z(z-1)(z-2)}$ around $z_0=1$ and the region is the whole complex plane. First I rewrote the functions using partial fractions and did some manipulation $f(z)=\frac{1}{z}\bigg(\frac{1}{1-z}-\frac{1}{2}\frac{1}{1-(\frac{z}{2})}\bigg)$ and then:
$$f(z)=\frac{1}{z}\bigg[\sum_{n=0}^{\infty}z^n - \sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}}\bigg]=\sum_{n=0}^{\infty}z^{n-1}\bigg(1-\frac{1}{2^{n+1}}\bigg)$$
The third one had me baffled.. $f(z)=\frac{e-e^z}{z(z-1)^2}$ around $z_0=1$ and in the region $0<|z-1|<1$. I tried the following:
$$f(z)=\frac{e}{z(z-1)^2}-\frac{1}{z(z-1)^2}\sum_{n=0}^{\infty}\frac{e(z-1)^n}{n!}=-\frac{e}{z(z-1)^2}\sum_{n=1}^{\infty}\frac{(z-1)^n}{n!}=-\frac{e}{z}\sum_{n=1}^{\infty}\frac{(z-1)^{n-2}}{n!}$$
Could any of these answers be considered correct? If not, what is the correct way to approach these problems?