Expand \(f(z) = {1\over z(z-1)}\) in both
- \(\abs{z} < 1\)
- \(\abs{z} > 1\)
Solution
\[ {1\over z(z-1)} = -{1\over z}{1 \over 1-z} = -{1\over z}\sum z^k .\] and \[ {1\over z(z-1)} = {1\over z^2(1 - {1\over z})} = {1\over z^2} \sum \qty{1\over z}^k .\]
Laurent series of $\frac{1}{z(z-1)}$ on $\abs z<1$ and on $\abs z>1$
E-BUVLS
- Area
- Complex Analysis
- Topics
- Laurent Series, Poles
- Status
- draft