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Find the laurent series expansion of $\frac{1}{z(z+5)}$ at z=0.

I found the solution for this question in MATLAB. The solution I got was $\frac{5}{z}+1$. I don't understand how the laurent series is this when I found out normally that for 0<|z|<5 the laurent series is $\sum_{n=0}^{n=\infty} \frac{(-1)^n z^{n-1}}{5^{n+1}}$

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    $\begingroup$ Is it possible you mis-typed into MATLAB? I ask because $$ \frac{5}{z}+1 = \frac{z+5}{z} \neq \frac{1}{z(z+5)}. $$ Your second series looks correct. $\endgroup$
    – WA Don
    May 13, 2020 at 6:23

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Since, when $|z|<5$,\begin{align}\frac1{z+5}&=\frac15\cdot\frac1{1+\frac z5}\\&=\frac15\sum_{n=0}^\infty\frac{(-1)^n}{5^n}z^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{5^{n+1}}z^n,\end{align}you have$$\frac1{z(z+5)}=\sum_{n=0}^\infty\frac{(-1)^n}{5^{n+1}}z^{n-1}=\sum_{n=-1}^\infty\frac{(-1)^{n+1}}{5^{n+2}}z^n.$$

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