Exercise · ZW-LAURENT-SERIES-007 · ★★☆ Intermediate
Let \(f(z)=\frac1{(z-1)(z-2)}\). On the annulus \(1<|z|<2\), what is the coefficient \(a_0\)?
Choose one option.
Show answer
- D. \(-\frac12\)
Explanation
By partial fractions \(f=\frac1{z-2}-\frac1{z-1}\). On \(1<|z|<2\), \(\frac1{z-2}=-\frac12\sum(z/2)^n\) contributes \(-\frac12\) to the constant term and \(-\frac1{z-1}=-\sum_{n\ge1}z^{-n}\) contributes nothing.
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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