Exercise · ZW-LAURENT-SERIES-002 · ★☆☆ Basic
Find the coefficient of \(z^{-1}\) in the Laurent expansion of \(z^2\sin(1/z)\) on \(0<|z|\).
Choose one option.
Show answer
- B. \(-\frac16\)
Explanation
Since \(\sin w=w-\frac{w^3}{6}+\cdots\), putting \(w=1/z\) gives \(z^2\sin(1/z)=z-\frac1{6z}+\cdots\). The coefficient of \(z^{-1}\) is \(-\frac16\).
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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