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# Expand $f(z)=\frac{1}{z(z-3)}$ in a Laurent series valid for the indicated annular domain. 1<|z-4|<4

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By the geometric series of $\frac{1}{1\pm z}$ in the domain $|z|<1$, we have

$\frac{1}{1+z}=1-z+z^2-z^3+\cdots$

$\frac{1}{1-z}=1+z+z^2+z^3+\cdots$

Consider the function $f(z)=\frac{1}{z(z-3)}$ in $1<|z-4|<4$. The domain $1<|z-4|<4$ equivalent to $|\frac{z-4}{4}|<1$ & $\frac{1}{|z-4|}<1$.

To expand $f(z)$ in terms of $\frac{1}{z-4}$ or $z-4$, rewrite the function $f(z)$ as follows:

\begin{align*} f(z)&=\frac{1}{z}\cdot \frac{1}{z-3}\\ &=\frac{1}{4+z-4}\cdot \frac{1}{1+z-4}\\ &=\frac{1}{4(z-4)}\frac{1}{1+\frac{z-4}{4}}\cdot \frac{1}{1+\frac{1}{z-4}}\\ \end{align*}

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