Find the Laurent series in the given domain:

Question

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

Answer 1

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 $$ \frac{1}{z}=\frac{1}{4}\cdot\frac{1}{1+\frac{z-4}{4}}. $$