I want to calculate the Laurent series of function : $$ f(z) = \frac{z}{(z-1)^3(z+1)}$$ in $z =1$.
I know that point $z=-1$ is the pole of this function. So I have to calculate Laurent series for $|z-1|<2$ and $|z-1|>2$. I tried this way:
$f(z)=\frac{z}{(z-1)^3(z+1)}= \frac{1}{8(z+1)}-\frac{1}{8(z-1)}+\frac{1}{4(z-1)^2}+\frac{1}{2(z-1)^3}$.
$f_1(z)=-\frac{1}{8(z-1)}=\frac{1}{16}\frac{1}{1-\frac{z+1}{2}}=\frac{1}{16}\sum(\frac{z+1}{2})^n$.
By analogy, $f_2(z)=\frac{1}{4}\frac{1}{(z-1)^2}=-\frac{1}{8}\sum(\frac{z+1}{2})^{2n}$ and $f_3(z)=\frac{1}{2}\frac{1}{(z-1)^3}=-\frac{1}{4}\sum(\frac{z+1}{2})^{3n}$. And finally, $$f(z)=\frac{1}{8(z+1)}+f_1(z)+f_2(z)+f_3(z)$$ Why I have to consider two cases $|z-1|<2$ and $|z-1|>2$ ? How is that different?
$$\frac{z}{z+1}=\frac{z}{2+(z-1)}=\frac{1}{2} \frac{z}{1+(z-1)/2} = \frac{(z-1)/2}{1+(z-1)/2}+\frac{1}{2}\cdot\frac{1}{1+(z-1)/2}$$
If $|z-1|<2$, then using the geometric series (convergent since $|z-1|/2<1$), you get that
$$\frac{z}{z+1}=\frac{(z-1)/2}{1+(z-1)/2}+\frac{1}{2}\cdot\frac{1}{1+(z-1)/2}=\sum_{n=1}^\infty \frac{1}{2^n}(z-1)^n +\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}(z-1)^n$$ Now $$\frac{z}{(z-1)^3)(z+1)}=\frac{1}{(z-1)^3}\cdot \frac{z}{z+1} = \sum_{n=1}^\infty \frac{1}{2^n}(z-1)^{n-3} +\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}(z-1)^{n-3}$$ and you can finish putting altogether in a single sum.
For $|z-1|>2$ you cannot use in this way the geometric series and need the next trick. $$\frac{z}{z+1}=\frac{z}{2+(z-1)}=\frac{1}{z-1} \cdot \frac{z}{1+2/(z-1)} = \frac{z}{z-1} \cdot \frac{1}{1+2/(z-1)}= \left(1+\frac{1}{z-1}\right)\frac{1}{1+2/(z-1)}.$$ Now, since $|z-1|>2$, the geometric series ((convergent since $2/|z-1|<1$)) says that $$\frac{1}{1+2/(z-1)}=\sum_{n=0}^\infty 2^n (z-1)^{-n}$$ and you probably can finish by yourself.