Exercise · ZW-LAURENT-SERIES-001 · ★☆☆ Basic

What is the coefficient of \(z^{-3}\) in the Laurent series of \(e^{1/z}\) on \(0<|z|\)?

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  • C. \(\frac16\)

Explanation

Substituting \(w=1/z\) in \(e^w=\sum w^k/k!\) gives \(\sum z^{-k}/k!\); for \(k=3\) the coefficient is \(1/3!=1/6\).

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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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