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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: enter image description here

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

Thanks in advance

Amaluena
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  • You are on the right track, but the only mistake is that your decomposition is wrong, since $1-\left(\frac{z-4}{4}\right) = \frac{8-z}{4}$. – Chee Han Dec 25 '17 at 11:58

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

Olivier Oloa
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  • Okay thanks, but what would this plus then mean for the geometric series, since in its most general form it is always with a minus, $\frac{1}{1-c}$ – Amaluena Dec 25 '17 at 12:01
  • @Amaluena Then you have $\frac1{1+c}=\sum_{j=0}^\infty(-1)^jc^j,, |c|<1,$ as noted in the handbook. – Olivier Oloa Dec 25 '17 at 12:02