Find the Laurent series in the given domain: I'm stuck on the following question: (it is from 'the fundamentals of complex analysis')
Find the Laurent series for $\frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$
Im working towards a geometric series in the following way:
$\frac{z+1}{z(z-4)^3}$=$\frac{1}{(z-4)^3}(1+\frac{1}{z})$
Then I tried rewriting the $\frac{1}{z}$ part since eventually for a geometric series we want something like $A*\frac{1}{1-\frac{z-4}{4}}$ since we know $|\frac{z-4}{4}|<1$
$\frac{1}{z}=-\frac{1}{4}*\frac{1}{1-\frac{z-4}{4}}=-\frac{1}{4}\sum_{j=0}^{\infty}(\frac{z-4}{4})^j$
So all together we'd obtain $\frac{1}{(z-4)^3}-\frac{1}{4}\sum_{j=0}^{\infty}(\frac{z-4}{4})^j$
However the solution manual says:

Can someone please help me to spot where I go wrong? Or are they just different answers that express the same?
Thanks in advance


Can someone please help me to spot where I go wrong?

There is a mistake when you write
$$
\frac{1}{z}=-\frac{1}{4}*\frac{1}{1-\frac{z-4}{4}}
$$ a correct version is rather
$$
\frac{1}{z}=\frac{1}{4}\cdot\frac{1}{1+\frac{z-4}{4}}.
$$
