Question

Expand:

`f(z) = (1)/(z(z - 2))`
for ;

`|z| > 2`
​

Answer 1

Expand f(z) into partial fractions:

`\frac1{z(z-2)} = \frac12 \left(\frac1{z-2} - \frac1z\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 (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 (2^n)/(z^(n+1)) - \frac1z\right)`

`\displaystyle f(z) = \frac12 \sum_(n=1)^\infty (2^n)/(z^(n+1))`

`\displaystyle f(z) = \sum_(n=1)^\infty (2^(n-1))/(z^(n+1))`

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