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Expand:
$$f(z) = \frac{1}{z(z - 2)}$$
for ;
$$|z| > 2$$

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Expand f(z) into partial fractions:

$$\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)$$

Recall that for |z| < 1, we have the power series

$$\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n$$

Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have

$$\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}$$

So the series expansion of f(z) for |z| > 2 is

$$\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)$$

$$\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}$$

$$\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}$$

$$\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}$$
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