Exercise · ZW-LAURENT-SERIES-004 · ★★☆ Intermediate
What is the Laurent series of \(\frac1{z-1}\) on the region \(|z|>1\)?
Choose one option.
Show answer
- A. \(\sum_{n=1}^{\infty}z^{-n}\)
Explanation
For \(|z|>1\): \(\frac1{z-1}=\frac1z\cdot\frac1{1-1/z}=\sum_{n\ge1}z^{-n}\). The option with positive powers corresponds to \(|z|<1\).
Review the concept: Laurent series
The Laurent series of a function \(f\) holomorphic on an annulus \(r<|z-z_0|<R\) is its expansion in integer powers, positive and negative, of \(z-z_0\), with coefficients \(a_n\).
\[f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n\]
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